postgresql exercises

PostgreSQL Übungen

Dies ist eine Zusammenstellung aller Fragen und Antworten zu Alisdair Owens Postgresql -Übungen. Denken Sie daran, dass Sie dazu führen, dass Sie dazu führen, dass Sie weiter als nur diesen Leitfaden durchfliegen. Wenn Sie also Postgresql -Übungen für einen Besuch zahlen.

Inhaltsverzeichnis

• Erste Schritte
  • Ich möchte mein eigenes Postgres -System verwenden
  • Schema
• Einfache SQL -Abfragen
  • ALLES ALLES VON EINEN TABEL ABEN
  • Rufen Sie bestimmte Spalten aus einer Tabelle ab
  • Steuerung, welche Zeilen abgerufen werden
  • Steuerung, welche Zeilen abgerufen werden, Teil 2
  • Grundlegende String -Suche
  • Übereinstimmung mit mehreren möglichen Werten
  • Die Ergebnisse in den Bucket eintreffen
  • Arbeiten mit Daten
  • Entfernen von Duplikaten und Bestellungsergebnisse
  • Kombination von Ergebnissen aus mehreren Abfragen
  • Einfache Aggregation
  • Mehr Aggregation
• Schließt sich und Unterabfragen
  • Abrufen Sie die Startzeiten der Mitgliederbuchungen ab
  • Arbeiten Sie die Anfangszeiten der Buchungen für Tennisplätze aus
  • Erstellen Sie eine Liste aller Mitglieder, die ein anderes Mitglied empfohlen haben
  • Erstellen Sie eine Liste aller Mitglieder zusammen mit ihren Empfehlungen
  • Erstellen Sie eine Liste aller Mitglieder, die einen Tennisplatz verwendet haben
  • Erstellen Sie eine Liste kostspieliger Buchungen
  • Erstellen Sie eine Liste aller Mitglieder zusammen mit ihrem Empfehlend
  • Erstellen Sie eine Liste kostspieliger Buchungen, indem Sie eine Unterabfrage verwenden
• Daten ändern
  • Fügen Sie einige Daten in eine Tabelle ein
  • Fügen Sie mehrere Datenzeilen in eine Tabelle ein
  • Fügen Sie berechnete Daten in eine Tabelle ein
  • Aktualisieren Sie einige vorhandene Daten
  • Aktualisieren Sie gleichzeitig mehrere Zeilen und Spalten
  • Aktualisieren Sie eine Zeile basierend auf dem Inhalt einer anderen Zeile
  • Alle Buchungen löschen
  • Löschen Sie ein Mitglied aus der C-Member-Tabelle
  • Basierend auf einer Unterabfrage löschen
• Aggregation
  • Zählen Sie die Anzahl der Einrichtungen
  • Zählen Sie die Anzahl der teuren Einrichtungen
  • Zählen Sie die Anzahl der Empfehlungen, die jedes Mitglied abgibt
  • Listen Sie die gesamten Slots auf, die pro Einrichtung gebucht wurden
  • Listen Sie die gesamten Slots pro Einrichtung in einem bestimmten Monat auf
  • Listen Sie die gesamten Slots auf, die pro Einrichtung und Monat gebucht wurden
  • Finden Sie die Anzahl von Mitgliedern, die mindestens eine Buchung vorgenommen haben
  • Listen Sie Einrichtungen mit mehr als 1000 Bucht auf
  • Finden Sie den Gesamtumsatz jeder Einrichtung
  • Finden Sie Einrichtungen mit einem Gesamtumsatz von weniger als 1000
  • Ausgabe der Facility -ID mit der höchsten Anzahl von Slots gebucht
  • Listen Sie die gesamten Slots auf, die pro Einrichtung und Monat gebucht wurden, Teil 2
  • Listen Sie die Gesamtstunden auf, die pro benannte Einrichtung gebucht wurden
  • Listen Sie die erste Buchung jedes Mitglieds nach dem 1. September 2012 auf
  • Erstellen Sie eine Liste der Mitgliedsnamen, wobei jede Zeile die Gesamtzahl der Mitglieder enthält
  • Erstellen Sie eine nummerierte Liste von Mitgliedern
  • Ausgabe der Facility -ID mit der höchsten Anzahl von Slots, die wieder gebucht wurden
  • Rang Mitglieder nach (abgerundeten) Stunden verwendet
  • Finden Sie die drei besten Einrichtungen zur Erzeugung von Einnahmen
  • Einrichtungen nach Wert klassifizieren
  • Berechnen Sie die Amortisationszeit für jede Einrichtung
  • Berechnen Sie einen rollenden Durchschnitt des Gesamtumsatzes
• Arbeiten mit Zeitstempeln
  • Erstellen Sie am 31. August 2012 einen Zeitstempel für 1 Uhr morgens
  • Subtrahieren Sie die Zeitstempel voneinander
  • Erstellen Sie eine Liste aller Daten im Oktober 2012
  • Holen Sie sich den Tag des Monats von einem Zeitstempel
  • Arbeiten Sie die Anzahl der Sekunden zwischen den Zeitstempeln aus
  • Arbeiten Sie die Anzahl der Tage in jedem Monat 2012 aus
  • Erarbeiten Sie die Anzahl der verbleibenden Tage im Monat
  • Arbeiten Sie die Endzeit der Buchungen aus
  • Gibt eine Anzahl von Buchungen für jeden Monat zurück
  • Erarbeiten Sie den Nutzungsanteil für jede Einrichtung bis Monat
• Stringoperationen
  • Formatieren Sie die Namen der Mitglieder
  • Finden Sie Einrichtungen mit einem Namenspräfix
  • Führen Sie eine unempfindliche Suche durch
  • Suchen Sie Telefonnummern mit Klammern
  • Pad -Reißverschlusscodes mit führenden Nullen
  • Zählen Sie die Anzahl der Mitglieder, deren Nachname mit jedem Buchstaben des Alphabets beginnt
  • Telefonnummern aufräumen
• Rekursive Abfragen
  • Finden Sie die Aufwärtsempfehlungskette für Mitglied ID 27
  • Finden Sie die Abwärts -Empfehlungskette für Mitglied ID 1
  • Erstellen Sie ein CTE, das die Aufwärtsbeschaffungskette für jedes Mitglied zurückgeben kann

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Erste Schritte

Es ist ziemlich einfach, mit den Übungen in Gang zu gehen: Alles, was Sie tun müssen, ist die Übungen zu öffnen, sich die Fragen anzusehen und sie zu beantworten!

Der Datensatz für diese Übungen gilt für einen neu erstellten Country Club mit einer Reihe von Mitgliedern, Einrichtungen wie Tennisplätzen und der Buchungsgeschichte für diese Einrichtungen. Unter anderem möchte der Club verstehen, wie er seine Informationen nutzen kann, um die Einrichtungsnutzung/-nachfrage zu analysieren. Bitte beachten Sie: Dieser Datensatz ist nur für die Unterstützung einer interessanten Auswahl von Übungen konzipiert, und das Datenbankschema ist in mehreren Aspekten fehlerhaft. Bitte nehmen Sie es nicht als Beispiel für ein gutes Design. Wir werden mit einem Blick auf den Mitgliedertisch beginnen:

 CREATE TABLE cd .members
(
    memid integer NOT NULL , 
    surname character varying ( 200 ) NOT NULL , 
    firstname character varying ( 200 ) NOT NULL , 
    address character varying ( 300 ) NOT NULL , 
    zipcode integer NOT NULL , 
    telephone character varying ( 20 ) NOT NULL , 
    recommendedby integer ,
    joindate timestamp not null ,
    CONSTRAINT members_pk PRIMARY KEY (memid),
    CONSTRAINT fk_members_recommendedby FOREIGN KEY (recommendedby)
        REFERENCES cd . members (memid) ON DELETE SET NULL
);

Jedes Mitglied verfügt über eine ID (nicht garantiert sequentiell), grundlegende Adressinformationen, einen Verweis auf das Mitglied, das sie empfohlen hat (falls vorhanden), und einen Zeitstempel, wenn es sich angeschlossen hat. Die Adressen im Datensatz sind vollständig (und unrealistisch) hergestellt.

 CREATE TABLE cd .facilities
(
    facid integer NOT NULL , 
    name character varying ( 100 ) NOT NULL , 
    membercost numeric NOT NULL , 
    guestcost numeric NOT NULL , 
    initialoutlay numeric NOT NULL , 
    monthlymaintenance numeric NOT NULL , 
    CONSTRAINT facilities_pk PRIMARY KEY (facid)
);

In der Tabelle der Einrichtungen werden alle buchbaren Einrichtungen aufgeführt, die der Country Club besitzt. Der Club speichert ID/Namensinformationen, die Kosten, um sowohl Mitglieder als auch Gäste zu buchen, die anfänglichen Kosten für den Bau der Einrichtung und schätzten die monatlichen Unterhaltskosten. Sie hoffen, diese Informationen zu nutzen, um zu verfolgen, wie sich jede Einrichtung finanziell lohnt.

 CREATE TABLE cd .bookings
(
    bookid integer NOT NULL , 
    facid integer NOT NULL , 
    memid integer NOT NULL , 
    starttime timestamp NOT NULL ,
    slots integer NOT NULL ,
    CONSTRAINT bookings_pk PRIMARY KEY (bookid),
    CONSTRAINT fk_bookings_facid FOREIGN KEY (facid) REFERENCES cd . facilities (facid),
    CONSTRAINT fk_bookings_memid FOREIGN KEY (memid) REFERENCES cd . members (memid)
);

Schließlich gibt es einen Tischverfolgungsbuch von Einrichtungen. Dies speichert den Facility -ID, das Mitglied, das die Buchung vorgenommen hat, zum Beginn der Buchung und wie viele halbstündige 'Slots' die Buchung gemacht wurde. Dieses eigenwillige Design wird bestimmte Abfragen schwieriger machen, sollte Ihnen jedoch einige interessante Herausforderungen stellen-sowie Sie auf den Schrecken vorbereiten, mit einigen realen Datenbanken zu arbeiten :-).

Okay, das sollte alle Informationen sein, die Sie benötigen. Sie können eine Abfragekategorie auswählen, die Sie aus dem obigen Menü ausprobieren können, oder von Anfang an alternativ beginnen.

Ich möchte mein eigenes Postgres -System verwenden

Kein Problem! Das Laufen ist nicht zu schwer. Erstens benötigen Sie eine Installation von PostgreSQL, die Sie von hier aus erhalten können. Sobald Sie es gestartet haben, laden Sie die SQL herunter.

Führen Sie schließlich psql -U <username> -f clubdata.sql -d postgres -x -q aus, um die Datenbank der 'Übungen' zu erstellen. das C -Gebietsschema)

Wenn Sie Fragen ausführen, finden Sie PSQL möglicherweise ein wenig klobig. In diesem Fall empfehle ich, PGADMIN oder die Eclipse -Datenbankentwicklungs -Tools auszuprobieren.

Schema

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Einfache SQL -Abfragen

Diese Kategorie befasst sich mit den Grundlagen von SQL. Es umfasst die Auswahl und wo Klauseln, Fallausdrücke, Gewerkschaften und einige andere Chancen und enden. Wenn Sie bereits in SQL ausgebildet sind, werden Sie diese Übungen wahrscheinlich ziemlich einfach finden. Wenn nicht, sollten Sie sie für einen guten Punkt finden, um für die schwierigeren Kategorien vor uns zu lernen!

Wenn Sie mit diesen Fragen zu kämpfen haben, empfehle ich dringend, SQL von Alan Beaulieu als prägnantes und gut geschriebenes Buch zu diesem Thema zu lernen. Wenn Sie an den Grundlagen von Datenbanksystemen interessiert sind (im Gegensatz zu genau der Verwendung), sollten Sie auch eine Einführung in Datenbanksysteme nach CJ -Datum untersuchen.

ALLES ALLES VON EINEN TABEL ABEN

Wie können Sie alle Informationen aus der Tabelle CD.Facities abrufen?

Erwartete Ergebnisse:

Gesicht	Name	Mitgliedskost	GuestCost	initialoutlay	Monatliche Wartung
0	Tennisplatz 1	5	25	10000	200
1	Tennisplatz 2	5	25	8000	200
2	Badminton Court	0	15.5	4000	50
3	Tischtennis	0	5	320	10
4	Massageraum 1	35	80	4000	3000
5	Massageraum 2	35	80	4000	3000
6	Kürbisgericht	3.5	17.5	5000	80
7	Snooker -Tisch	0	5	450	15
8	Billardtisch	0	5	400	15

Antwort:

 select * from cd . facilities ;          

Die SELECT ist der grundlegende Startblock für Abfragen, die Informationen aus der Datenbank herauslesen. Eine minimale Auswahlanweisung besteht im Allgemeinen aus select [some set of columns] from [some table or group of tables] .

In diesem Fall möchten wir alle Informationen aus der Tabelle der Einrichtungen. Der Abschnitt ist einfach - wir müssen nur die cd.facilities -Tabelle angeben. 'CD' ist das Schema der Tabelle - ein Begriff, der für eine logische Gruppierung verwandter Informationen in der Datenbank verwendet wird.

Als nächstes müssen wir angeben, dass wir alle Spalten wollen. Praktischerweise gibt es eine Abkürzung für "alle Säulen" - *. Wir können dies verwenden, anstatt alle Spaltennamen mühsam anzugeben.

Rufen Sie bestimmte Spalten aus einer Tabelle ab

Sie möchten eine Liste aller Einrichtungen und ihrer Kosten für Mitglieder ausdrucken. Wie würden Sie eine Liste nur von den Namen und Kosten für Einrichtungen abrufen?

Erwartete Ergebnisse:

Name	Mitgliedskost
Tennisplatz 1	5
Tennisplatz 2	5
Badminton Court	0
Tischtennis	0
Massageraum 1	35
Massageraum 2	35
Kürbisgericht	3.5
Snooker -Tisch	0
Billardtisch	0

Antwort:

 select name, membercost from cd . facilities ;          

Für diese Frage müssen wir die gewünschten Spalten angeben. Wir können dies mit einer einfachen, von Kommas mit Delimited-Liste der in der Auswahlanweisung angegebenen Spaltennamen tun. Alles in der Datenbank wird die in der Ab -Klausel verfügbaren Spalten betrachtet und die von uns gefragten zurückgeben, wie unten dargestellt

Im Allgemeinen gilt für Nicht-Throwaway-Abfragen als wünschenswert, die Namen der gewünschten Spalten in Ihren Abfragen zu geben, anstatt *zu verwenden. Dies liegt daran, dass Ihre Bewerbung möglicherweise nicht in der Lage ist, damit fertig zu werden, wenn mehr Spalten in die Tabelle hinzugefügt werden.

Steuerung, welche Zeilen abgerufen werden

Wie können Sie eine Liste von Einrichtungen erstellen, die den Mitgliedern eine Gebühr erheben?

Erwartete Ergebnisse:

Gesicht	Name	Mitgliedskost	GuestCost	initialoutlay	Monatliche Wartung
0	Tennisplatz 1	5	25	10000	200
1	Tennisplatz 2	5	25	8000	200
4	Massageraum 1	35	80	4000	3000
5	Massageraum 2	35	80	4000	3000
6	Kürbisgericht	3.5	17.5	5000	80

Antwort:

 select * from cd . facilities where membercost > 0 ;          

Die FROM -Klausel wird verwendet, um eine Reihe von Kandidatenreihen aufzubauen, um Ergebnisse zu lesen. In unseren bisherigen Beispielen war diese Reihe von Zeilen einfach der Inhalt einer Tabelle. In Zukunft werden wir das Zusammenschluss untersuchen, was es uns ermöglicht, viel interessantere Kandidaten zu schaffen.

Sobald wir unsere Kandidatenreihen aufgebaut haben, ermöglicht die WHERE , für die Zeilen, an denen wir interessiert sind, zu filtern - in diesem Fall diejenigen mit einem Mitglied von mehr als Null. Wie Sie in späteren Übungen sehen werden, WHERE Klauseln mehrere Komponenten in Kombination mit einer WHERE Logik haben können, ist es möglich, nach Einrichtungen mit einer Kosten von mehr als 0 und weniger als 10 zu suchen.

Steuerung, welche Zeilen abgerufen werden, Teil 2

Wie können Sie eine Liste von Einrichtungen erstellen, die den Mitgliedern eine Gebühr erheben, und diese Gebühr beträgt weniger als 1/50 der monatlichen Wartungskosten? Geben Sie den Namen, den Namen der Einrichtungen, die Mitgliederkosten und die monatliche Wartung der fraglichen Einrichtungen zurück.

Erwartete Ergebnisse:

Gesicht	Name	Mitgliedskost	Monatliche Wartung
4	Massageraum 1	35	3000
5	Massageraum 2	35	3000

Antwort:

 select facid, name, membercost, monthlymaintenance 
	from cd . facilities 
	where 
		membercost > 0 and 
		(membercost < monthlymaintenance / 50 . 0 );          

In der WHERE -Klausel können wir die Zeilen filtern, an denen wir interessiert sind - in diesem Fall diejenigen mit einem Mitgliedskost von mehr als Null und weniger als 1/50 der monatlichen Wartungskosten. Wie Sie sehen können, sind die Massageberäte dank der Personalkosten sehr teuer zu laufen!

Wenn wir zwei oder mehr Bedingungen testen wollen, verwenden wir sie AND kombinieren sie. Wir können, wie Sie vielleicht erwarten, verwenden OR testen können, ob ein Paar von Bedingungen wahr ist.

Möglicherweise haben Sie festgestellt, dass dies unsere erste Abfrage ist, die eine WHERE -Klausel mit Auswahl bestimmter Spalten kombiniert. Sie können im Bild unten sehen: Die Schnittstelle der ausgewählten Spalten und der ausgewählten Zeilen gibt uns die Daten zur Rückgabe. Dies scheint jetzt nicht zu interessant zu sein, aber wenn wir komplexere Operationen wie Joins später hinzufügen, werden Sie die einfache Eleganz dieses Verhaltens sehen.

Grundlegende String -Suche

Wie können Sie eine Liste aller Einrichtungen mit dem Wort "Tennis" in ihrem Namen erstellen?

Erwartete Ergebnisse:

Gesicht	Name	Mitgliedskost	GuestCost	initialoutlay	Monatliche Wartung
0	Tennisplatz 1	5	25	10000	200
1	Tennisplatz 2	5	25	8000	200
3	Tischtennis	0	5	320	10

Antwort:

 select *
	from cd . facilities 
	where 
		name like ' %Tennis% ' ;          

SQLs LIKE Operator bietet einfache Muster -Anpassungen für Saiten. Es ist ziemlich allgemein implementiert und ist schön und einfach zu bedienen - es nimmt nur eine Zeichenfolge mit dem % -Scharn an, das zu einer Zeichenfolge passt, und _ für ein einzelnes Zeichen. In diesem Fall suchen wir nach Namen, die das Wort "Tennis" enthalten, sodass ein % auf beide Seiten in die Rechnung passt.

Es gibt andere Möglichkeiten, diese Aufgabe zu erfüllen: Postgres unterstützt beispielsweise regelmäßige Ausdrücke mit dem ~ Operator. Verwenden Sie alles, was Sie sich wohl fühlen, aber sind Sie sich bewusst, dass der LIKE Betreiber zwischen den Systemen viel tragbarer ist.

Übereinstimmung mit mehreren möglichen Werten

Wie können Sie die Details von Einrichtungen mit ID 1 und 5 abrufen? Versuchen Sie es, es ohne den oder den Bediener zu tun.

Erwartete Ergebnisse:

Gesicht	Name	Mitgliedskost	GuestCost	initialoutlay	Monatliche Wartung
1	Tennisplatz 2	5	25	8000	200
5	Massageraum 2	35	80	4000	3000

Antwort:

 select *
	from cd . facilities 
	where 
		facid in ( 1 , 5 );          

Die offensichtliche Antwort auf diese Frage ist, eine WHERE -Klausel zu verwenden, die so aussieht, als where facid = 1 or facid = 5 . Eine Alternative, die bei einer großen Anzahl möglicher Übereinstimmungen einfacher ist, ist der IN . Der IN -Operator nimmt eine Liste möglicher Werte an und entspricht ihnen gegen (in diesem Fall) der Gesichtspunkte. Wenn einer der Werte übereinstimmt, gilt die WHERE -Klausel für diese Zeile, und die Zeile wird zurückgegeben.

Der IN -Operator ist ein guter früher Demonstrator der Eleganz des relationalen Modells. Das Argument, das es nimmt, ist nicht nur eine Liste von Werten - es ist tatsächlich eine Tabelle mit einer einzelnen Spalte. Da Abfragen auch Tabellen zurückgeben, können Sie diese Ergebnisse in einen IN -Operator einfügen, wenn Sie eine Abfrage erstellen, die eine einzelne Spalte zurückgibt. Ein Spielzeugbeispiel geben:

 select * 
	from cd . facilities
	where
		facid in (
			select facid from cd . facilities
			);

Dieses Beispiel entspricht funktional der Auswahl aller Einrichtungen, zeigt jedoch, wie Sie die Ergebnisse einer Abfrage in eine andere einfügen. Die innere Abfrage wird als Unterabfrage bezeichnet.

Die Ergebnisse in den Bucket eintreffen

Wie können Sie eine Liste von Einrichtungen erstellen, wobei jedes als "billig" oder "teuer" bezeichnet wird, je nachdem, ob ihre monatlichen Wartungskosten mehr als 100 US -Dollar beträgt? Geben Sie den Namen und die monatliche Wartung der fraglichen Einrichtungen zurück.

Erwartete Ergebnisse:

Name	kosten
Tennisplatz 1	teuer
Tennisplatz 2	teuer
Badminton Court	billig
Tischtennis	billig
Massageraum 1	teuer
Massageraum 2	teuer
Kürbisgericht	billig
Snooker -Tisch	billig
Billardtisch	billig

Antwort:

 select name, 
	case when (monthlymaintenance > 100 ) then
		' expensive '
	else
		' cheap '
	end as cost
	from cd . facilities ;          

Diese Übung enthält einige neue Konzepte. Die erste ist die Tatsache, dass wir im Bereich der Abfrage zwischen SELECT und FROM berechnet werden. Früher haben wir dies nur verwendet, um Spalten auszuwählen, die wir zurückkehren möchten, aber Sie können hier alles einfügen, das ein einziges Ergebnis pro zurückgegebener Zeile erzeugt - einschließlich Unterabfragen.

Das zweite neue Konzept ist die CASE selbst. CASE ist effektiv, wenn/Switch -Anweisungen in anderen Sprachen mit einem Formular wie in der Abfrage gezeigt. Um eine "mittelmäßige" Option hinzuzufügen, würden wir einfach einen weiteren einfügen when...then der Abschnitt.

Schließlich gibt es den AS Operator. Dies wird einfach verwendet, um Spalten oder Ausdrücke zu kennzeichnen, damit sie besser angezeigt werden oder sie leichter zu verweisen, wenn sie als Teil einer Unterabfrage verwendet werden.

Arbeiten mit Daten

Wie können Sie eine Liste von Mitgliedern erstellen, die nach Anfang September 2012 beigetreten sind? Geben Sie den Memid, den Nachnamen, den FirstName und gemeinsam der fraglichen Mitglieder zurück.

Erwartete Ergebnisse:

Memid	Nachname	Erstname	zusammen
24	Sarwin	Ramnaresh	2012-09-01 08:44:42
26	Jones	Douglas	2012-09-02 18:43:05
27	Rumney	Henrietta	2012-09-05 08:42:35
28	Farrell	David	2012-09-15 08:22:05
29	Worthington-Smyth	Henry	2012-09-17 12:27:15
30	Aufgabe	Millicent	2012-09-18 19:04:01
33	Tupperware	Hyazinthe	2012-09-18 19:32:05
35	Jagd	John	2012-09-19 11:32:45
36	Crumpet	Erica	2012-09-22 08:36:38
37	Schmied	Darren	2012-09-26 18:08:45

Antwort:

 select memid, surname, firstname, joindate 
	from cd . members
	where joindate >= ' 2012-09-01 ' ;          

Dies ist unser erster Blick auf SQL Timestemps. Sie sind in absteigender Größenordnung formatiert: YYYY-MM-DD HH:MM:SS.nnnnnn . Wir können sie genauso vergleichen wie ein Unix -Zeitstempel, obwohl es etwas mehr involviert (und mächtig!) Es ist etwas mehr involviert (und mächtig!). In diesem Fall haben wir gerade den Datumsanteil des Zeitstempels angegeben. Dies wird automatisch von Postgres in den vollständigen Timestamp 2012-09-01 00:00:00 gegossen.

Entfernen von Duplikaten und Bestellungsergebnisse

Wie können Sie eine bestellte Liste der ersten 10 Nachnamen in der Mitgliedertabelle erstellen? Die Liste darf keine Duplikate enthalten.

Erwartete Ergebnisse:

Nachname
Bader
Bäcker
Stand
Butters
Coplin
Crumpet
Wagen
Farrell
GAST
Genting

Antwort:

 select distinct surname 
	from cd . members
order by surname
limit 10 ;          

Hier gibt es drei neue Konzepte, aber sie sind alle ziemlich einfach.

• Das Angeben DISTINCT nach SELECT wird doppelte Zeilen aus dem Ergebnissatz entfernt. Beachten Sie, dass dies für Zeilen gilt: Wenn Zeile A mehrere Spalten enthält, ist Zeile B nur gleich, wenn die Werte in allen Spalten gleich sind. Verwenden Sie in der Regel nicht DISTINCT , dass Sie sich nicht milles Weise befinden-es ist nicht frei, Duplikate aus großen Abfrageergebnis-Sets zu entfernen.
• Angabe ORDER BY ( FROM und WHERE Klauseln gegen Ende der Abfrage) ermöglicht, die Ergebnisse durch eine Spalte oder eine Reihe von Spalten (Comma getrennt) zu bestellen.
• Mit dem Schlüsselwort LIMIT können Sie die Anzahl der abgerufenen Ergebnisse begrenzen. Dies ist nützlich, um Ergebnisse jeweils eine Seite zu erhalten, und kann mit dem OFFSET -Keyword kombiniert werden, um folgende Seiten zu erhalten. Dies ist der gleiche Ansatz, der von MySQL verwendet wird und sehr bequem ist - Sie können leider feststellen, dass dieser Prozess in anderen DBs etwas komplizierter ist.

Kombination von Ergebnissen aus mehreren Abfragen

Sie möchten aus irgendeinem Grund eine kombinierte Liste aller Nachnamen und aller Einrichtungsnamen. Ja, dies ist ein erfundenes Beispiel :-). Erstellen Sie diese Liste!

Erwartete Ergebnisse:

Nachname
Tennisplatz 2
Worthington-Smyth
Badminton Court
Pinker
Wagen
Bader
Mackenzie
Crumpet
Massageraum 1
Kürbisgericht

Antwort:

 select surname 
	from cd . members
union
select name
	from cd . facilities ;          

Der UNION macht das, was Sie erwarten könnten: kombiniert die Ergebnisse von zwei SQL -Abfragen in eine einzelne Tabelle. Die Einschränkung besteht darin, dass beide Ergebnisse der beiden Abfragen die gleiche Anzahl von Spalten und kompatiblen Datentypen haben müssen.

UNION entfernt doppelte Reihen, während UNION ALL nicht. Verwenden Sie UNION ALL standardmäßig, es sei denn, Sie kümmern sich um doppelte Ergebnisse.

Einfache Aggregation

Sie möchten das Anmeldedatum Ihres letzten Mitglieds erhalten. Wie können Sie diese Informationen abrufen?

Erwartete Ergebnisse:

letzte
2012-09-26 18:08:45

Antwort:

 select max (joindate) as latest
	from cd . members ;          

Dies ist unser erster Ausflug in die Gesamtfunktionen von SQL. Sie werden verwendet, um Informationen über ganze Gruppen von Zeilen zu extrahieren und es uns zu ermöglichen, leicht Fragen zu stellen wie:

• Was ist die teuerste Einrichtung, um monatlich aufrechtzuerhalten?
• Wer hat die neuen Mitglieder empfohlen?
• Wie viel Zeit hat jedes Mitglied in unseren Einrichtungen verbracht?

Die MAX -Aggregat -Funktion hier ist sehr einfach: Sie empfängt alle möglichen Werte für Joindate und gibt den größten aus. Es gibt viel mehr Kraft, um Funktionen zu aggregieren, auf die Sie in zukünftigen Übungen stoßen werden.

Mehr Aggregation

Sie möchten den ersten und Nachnamen der letzten Mitglieder bekommen, die sich angemeldet haben - nicht nur das Datum. Wie kannst du das machen?

Erwartete Ergebnisse:

Erstname	Nachname	zusammen
Darren	Schmied	2012-09-26 18:08:45

Antwort:

 select firstname, surname, joindate
	from cd . members
	where joindate = 
		( select max (joindate) 
			from cd . members );          

In dem vorgeschlagenen Ansatz oben verwenden Sie eine Unterabfrage , um herauszufinden, was die jüngste Joindate ist. Diese Unterabfrage gibt eine skalare Tabelle zurück - dh eine Tabelle mit einer einzelnen Spalte und einer einzelnen Zeile. Da wir nur einen einzelnen Wert haben, können wir die Unterabfrage überall ersetzen, wo wir möglicherweise einen einzigen konstanten Wert setzen. In diesem Fall verwenden wir es, um die WHERE einer Abfrage zu vervollständigen, um ein bestimmtes Mitglied zu finden.

Sie könnten hoffen, dass Sie so etwas wie unten tun können:

 select firstname, surname, max (joindate)
        from cd . members

Leider funktioniert das nicht. Die MAX -Funktion schränkt die Zeilen nicht so ein, wie die WHERE -Klausel - sie nimmt einfach eine Reihe von Werten ein und gibt den größten zurück. Die Datenbank bleibt dann gefragt, wie man eine lange Liste von Namen mit dem einzigen Join -Datum, der aus der MAX -Funktion stammt, kombiniert und fehlschlägt. Stattdessen müssen Sie sagen: "Finden Sie mir die Reihe (en), die ein Verbindungdatum haben, das dem maximalen Join -Datum entspricht."

Wie der Hinweis erwähnt, gibt es andere Möglichkeiten, diesen Job zu erledigen - ein Beispiel ist unten. Bei diesem Ansatz bestellen wir einfach ausdrücklich heraus, was das letzte Datum des letzten zusammengehaltenen Datums in der absteigenden Reihenfolge des Verbindungsdatums bestellen und den ersten abnehmen. Beachten Sie, dass dieser Ansatz nicht die äußerst unwahrscheinliche Eventualität von zwei Personen abdeckt, die genau zur gleichen Zeit beigetreten sind :-).

 select firstname, surname, joindate
	from cd . members
order by joindate desc
limit 1 ;

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Schließt sich und Unterabfragen

Diese Kategorie befasst sich hauptsächlich mit einem grundlegenden Konzept in relationalen Datenbanksystemen: Joining. Mit Join können Sie verwandte Informationen aus mehreren Tabellen kombinieren, um eine Frage zu beantworten. Dies ist nicht nur vorteilhaft für die einfache Abfrage: Ein Mangel an Join -Fähigkeit fördert die Denormalisierung von Daten, wodurch die Komplexität der Innenkonsistent angehalten wird.

Dieses Thema deckt innere, äußere und Selbstverbindungen ab und verbringt ein wenig Zeit mit Unterabfragen (Abfragen innerhalb von Abfragen). Wenn Sie mit diesen Fragen zu kämpfen haben, empfehle ich dringend, SQL von Alan Beaulieu als prägnantes und gut geschriebenes Buch zu diesem Thema zu lernen.

Abrufen Sie die Startzeiten der Mitgliederbuchungen ab

Wie können Sie eine Liste der Anfangszeiten für Buchungen von Mitgliedern namens "David Farrell" erstellen?

Erwartete Ergebnisse:

Startzeit
2012-09-18 09:00:00
2012-09-18 17:30:00
2012-09-18 13:30:00
2012-09-18 20:00:00
2012-09-19 09:30:00
2012-09-19 15:00:00
2012-09-19 12:00:00
2012-09-20 15:30:00
2012-09-20 11:30:00
2012-09-20 14:00:00

Antwort:

 select bks . starttime 
	from 
		cd . bookings bks
		inner join cd . members mems
			on mems . memid = bks . memid
	where 
		mems . firstname = ' David ' 
		and mems . surname = ' Farrell ' ;          

Die am häufigsten verwendete Art von Join ist der INNER JOIN . Dies kombiniert zwei Tabellen basierend auf einem Join -Ausdruck - in diesem Fall suchen wir für jede Mitglieds -ID in der Mitgliedertabelle nach passenden Werten in der Buchungstabelle. Wo wir eine Übereinstimmung finden, wird eine Zeile, die die Werte für jede Tabelle kombiniert, zurückgegeben. Beachten Sie, dass wir jeder Tabelle einen Alias ​​(Bks und Mems) gegeben haben. Dies wird aus zwei Gründen verwendet: Erstens ist es bequem, und zweitens könnten wir uns mehrmals mit derselben Tabelle verbinden, sodass wir zwischen den Spalten von jeder verschiedenen Zeit, in der die Tabelle verbunden war, unterscheidet.

Ignorieren wir unsere Auswahl und wo vorerst Klauseln, und konzentrieren uns auf das, was die FROM hervorbringt. In all unseren früheren Beispielen war FROM gerade eine einfache Tabelle. Was ist es jetzt? Noch ein Tisch! Diesmal wird es als Komposit aus Buchungen und Mitgliedern produziert. Sie können eine Teilmenge der Ausgabe des Join unten sehen:

Für jedes Mitglied in der Mitgliedertabelle hat der Join alle passenden Mitglieder -IDs in der Buchungstabelle gefunden. Für jedes Match wird dann eine Zeile produziert, die die Zeile aus der Mitgliedertabelle und die Zeile aus der Buchungstabelle kombiniert.

Offensichtlich sind dies zu viele Informationen selbst, und jede nützliche Frage wird sie filtern. In unserer Abfrage verwenden wir den Beginn der SELECT -Klausel, um Spalten auszuwählen, und die WHERE -Klausel, um Zeilen auszuwählen, wie unten dargestellt:

Das ist alles, was wir brauchen, um Davids Buchungen zu finden! Im Allgemeinen ermutige ich Sie, sich daran zu erinnern, dass die Ausgabe der FROM -Klausel im Wesentlichen eine große Tabelle ist, aus der Sie dann Informationen herausfiltern. Das mag ineffizient klingen - aber keine Sorge, unter den Abdeckungen wird sich die DB viel intelligenter verhalten :-).

Ein letzter Anmerkung: Es gibt zwei verschiedene Syntaxen für innere Zusammenhänge. Ich habe dir das gezeigt, das ich bevorzuge, das ich mit anderen Join -Typen besser übereinstimme. Normalerweise sehen Sie eine andere Syntax, siehe unten:

 select bks . starttime
        from
                cd . bookings bks,
                cd . members mems
        where
                mems . firstname = ' David '
                and mems . surname = ' Farrell '
                and mems . memid = bks . memid ;

Dies ist funktional genau das gleiche wie die genehmigte Antwort. Wenn Sie sich mit dieser Syntax wohler fühlen, können Sie sie gerne verwenden!

Arbeiten Sie die Anfangszeiten der Buchungen für Tennisplätze aus

Wie können Sie eine Liste der Startzeiten für Buchungen für Tennisplätze für das Datum "2012-09-21" erstellen? Geben Sie eine Liste der Pairings für Startzeit und Einrichtungsname zurück, die bis zu diesem Zeitpunkt bestellt wurden.

Erwartete Ergebnisse:

Start	Name
2012-09-21 08:00:00	Tennisplatz 1
2012-09-21 08:00:00	Tennisplatz 2
2012-09-21 09:30:00	Tennisplatz 1
2012-09-21 10:00:00	Tennisplatz 2
2012-09-21 11:30:00	Tennisplatz 2
2012-09-21 12:00:00	Tennisplatz 1
2012-09-21 13:30:00	Tennisplatz 1
2012-09-21 14:00:00	Tennisplatz 2
2012-09-21 15:30:00	Tennisplatz 1
2012-09-21 16:00:00	Tennisplatz 2
2012-09-21 17:00:00	Tennisplatz 1
2012-09-21 18:00:00	Tennisplatz 2

Antwort:

 select bks . starttime as start, facs . name as name
	from 
		cd . facilities facs
		inner join cd . bookings bks
			on facs . facid = bks . facid
	where 
		facs . facid in ( 0 , 1 ) and
		bks . starttime >= ' 2012-09-21 ' and
		bks . starttime < ' 2012-09-22 '
order by bks . starttime ;          

Dies ist eine weitere INNER JOIN Abfrage, obwohl es ein bisschen mehr Komplexität hat! Die FROM einem Teil der Abfrage ist einfach - wir schließen uns einfach zusammen mit Einrichtungen und Buchungen zusammen auf der Gesichtspunkte an. Dies erzeugt eine Tabelle, in der wir für jede Zeile in Buchungen detaillierte Informationen über die gebuchte Einrichtung beigefügt haben.

Auf die WHERE der Komponente der Abfrage. Die Überprüfungen der Starttime sind ziemlich selbsterklärend - wir stellen sicher, dass alle Buchungen zwischen den angegebenen Daten beginnen. Da wir nur an Tennisplätzen interessiert sind, verwenden wir auch den IN -Betreiber, um das Datenbanksystem zu sagen, dass wir uns nur die IDs 0 oder 1 zurückgeben sollen - die IDs der Gerichte. Es gibt andere Möglichkeiten, dies auszudrücken: Wir hätten verwenden können where facs.facid = 0 or facs.facid = 1 oder sogar where facs.name like 'Tennis%' verwendet werden.

Der Rest ist ziemlich einfach: Wir SELECT die Spalten aus, an denen wir interessiert sind, und ORDER BY zur Startzeit.

Erstellen Sie eine Liste aller Mitglieder, die ein anderes Mitglied empfohlen haben

Wie können Sie eine Liste aller Mitglieder ausgeben, die ein anderes Mitglied empfohlen haben? Stellen Sie sicher, dass in der Liste keine Duplikate vorhanden sind und dass die Ergebnisse von (Nachname, FirstName) bestellt werden.

Erwartete Ergebnisse:

Erstname	Nachname
Florenz	Bader
Timothy	Bäcker
Gerald	Butters
Jemima	Farrell
Matthew	Genting
David	Jones
Janice	JOPLETTE
Millicent	Aufgabe
Tim	Rownam
Darren	Schmied
Tracy	Schmied
Überlegen	Stibbons
Burton	Tracy

Antwort:

 select distinct recs . firstname as firstname, recs . surname as surname
	from 
		cd . members mems
		inner join cd . members recs
			on recs . memid = mems . recommendedby
order by surname, firstname;          

Hier ist ein Konzept, das manche Menschen verwirrend empfinden: Sie können sich einem Tisch an sich selbst anschließen! Dies ist sehr nützlich, wenn Sie Spalten haben, die Daten in derselben Tabelle verweisen, wie wir es mit empfohlenen von CD.MEMBERS empfohlen haben.

Wenn Sie Probleme haben, dies zu visualisieren, denken Sie daran, dass dies genauso wie jeder andere innere Join funktioniert. Unser Join nimmt jede Zeile in Mitgliedern mit einem empfohlenen Wert ein und schaut sich erneut in den Mitgliedern für die Zeile mit einer passenden Mitglieder -ID an. Anschließend erzeugt eine Ausgabezeile, die die beiden Mitgliedereinträge kombiniert. Dies sieht aus wie das folgende Diagramm:

Beachten Sie, dass wir zwar zwei "Nachname" -Spalten im Ausgangssatz haben, sie jedoch durch ihre Tischaliase unterschieden werden können. Sobald wir die gewünschten Spalten ausgewählt haben, verwenden wir einfach DISTINCT , um sicherzustellen, dass es keine Duplikate gibt.

Erstellen Sie eine Liste aller Mitglieder zusammen mit ihren Empfehlungen

Wie können Sie eine Liste aller Mitglieder ausgeben, einschließlich der Person, die sie empfohlen hat (falls vorhanden)? Stellen Sie sicher, dass die Ergebnisse von (Nachname, FirstName) bestellt werden.

Erwartete Ergebnisse:

Memfname	Memsname	Recfnname	Recsname
Florenz	Bader	Überlegen	Stibbons
Anne	Bäcker	Überlegen	Stibbons
Timothy	Bäcker	Jemima	Farrell
Tim	Stand	Tim	Rownam
Gerald	Butters	Darren	Schmied
Joan	Coplin	Timothy	Bäcker
Erica	Crumpet	Tracy	Schmied
Nancy	Wagen	Janice	JOPLETTE
David	Farrell
Jemima	Farrell
GAST	GAST
Matthew	Genting	Gerald	Butters
John	Jagd	Millicent	Aufgabe
David	Jones	Janice	JOPLETTE
Douglas	Jones	David	Jones
Janice	JOPLETTE	Darren	Schmied
Anna	Mackenzie	Darren	Schmied
Charles	Owen	Darren	Schmied
David	Pinker	Jemima	Farrell
Millicent	Aufgabe	Tracy	Schmied
Tim	Rownam
Henrietta	Rumney	Matthew	Genting
Ramnaresh	Sarwin	Florenz	Bader
Darren	Schmied
Darren	Schmied
Jack	Schmied	Darren	Schmied
Tracy	Schmied
Überlegen	Stibbons	Burton	Tracy
Burton	Tracy
Hyazinthe	Tupperware
Henry	Worthington-Smyth	Tracy	Schmied

Antwort:

 select mems . firstname as memfname, mems . surname as memsname, recs . firstname as recfname, recs . surname as recsname
	from 
		cd . members mems
		left outer join cd . members recs
			on recs . memid = mems . recommendedby
order by memsname, memfname;          

Lasst uns ein weiteres neues Konzept vorstellen: den LEFT OUTER JOIN . Diese werden am besten durch die Art und Weise erklärt, wie sie sich von den inneren Anschlüssen unterscheiden. Innenverbindungen nehmen einen linken und einen rechten Tisch und suchen nach passenden Zeilen, die auf einer Join -Bedingung ( ON ) basieren. Wenn die Bedingung erfüllt ist, wird eine zusammengeführte Reihe erzeugt. Ein LEFT OUTER JOIN funktioniert ähnlich, außer dass eine bestimmte Zeile in der linken Handtabelle nichts übereinstimmt, sondern immer noch eine Ausgangszeile erzeugt. Diese Ausgangszeile besteht aus der linken Tischreihe und einer Reihe von NULLS anstelle der rechten Tabellenzeile.

Dies ist in Situationen wie dieser Frage nützlich, in denen wir mit optionalen Daten Ausgaben erzeugen möchten. Wir wollen die Namen aller Mitglieder und den Namen ihres Empfehlends , wenn diese Person existiert . Sie können das mit einem inneren Join nicht richtig ausdrücken.

Wie Sie vielleicht vermutet haben, gibt es auch andere äußere Zusammenhänge. Der RIGHT OUTER JOIN ähnelt dem LEFT OUTER JOIN , außer dass die linke Seite des Ausdrucks diejenige ist, die die optionalen Daten enthält. Der selten verwendete FULL OUTER JOIN behandelt beide Seiten des Ausdrucks als optional.

Erstellen Sie eine Liste aller Mitglieder, die einen Tennisplatz verwendet haben

Wie können Sie eine Liste aller Mitglieder erstellen, die ein Tennisplatz verwendet haben? Fügen Sie in Ihre Ausgabe den Namen des Gerichts und den Namen des als einzelnen Spalte formatierten Mitglieds ein. Stellen Sie keine doppelten Daten sicher und bestellen Sie nach dem Mitgliedsnamen.

Erwartete Ergebnisse:

Mitglied	Einrichtung
Anne Baker	Tennisplatz 2
Anne Baker	Tennisplatz 1
Burton Tracy	Tennisplatz 2
Burton Tracy	Tennisplatz 1
Charles Owen	Tennisplatz 2
Charles Owen	Tennisplatz 1
Darren Smith	Tennisplatz 2
David Farrell	Tennisplatz 2
David Farrell	Tennisplatz 1
David Jones	Tennisplatz 1
David Jones	Tennisplatz 2
David Pinker	Tennisplatz 1
Douglas Jones	Tennisplatz 1
Erica Crumpet	Tennisplatz 1
Florence Bader	Tennisplatz 1
Florence Bader	Tennisplatz 2
Gast	Tennisplatz 2
Gast	Tennisplatz 1
Gerald Butters	Tennisplatz 1
Gerald Butters	Tennisplatz 2
Henrietta Rumney	Tennisplatz 2
Jack Smith	Tennisplatz 1
Jack Smith	Tennisplatz 2
Janice Joplette	Tennisplatz 1
Janice Joplette	Tennisplatz 2
Jemima Farrell	Tennisplatz 2
Jemima Farrell	Tennisplatz 1
Joan Coplin	Tennisplatz 1
John Hunt	Tennisplatz 1
John Hunt	Tennisplatz 2
Matthew Genting	Tennisplatz 1
Millicent -Zuständigkeitsbereich	Tennisplatz 2
Nancy Dare	Tennisplatz 2
Nancy Dare	Tennisplatz 1
Überlegen Sie Stibbons	Tennisplatz 2
Überlegen Sie Stibbons	Tennisplatz 1
Ramnaresh Sarwin	Tennisplatz 2
Ramnaresh Sarwin	Tennisplatz 1
Tim Boothe	Tennisplatz 1
Tim Boothe	Tennisplatz 2
Tim Rownam	Tennisplatz 1
Tim Rownam	Tennisplatz 2
Timothy Baker	Tennisplatz 2
Timothy Baker	Tennisplatz 1
Tracy Smith	Tennisplatz 2
Tracy Smith	Tennisplatz 1

Antwort:

 select distinct mems . firstname || ' ' || mems . surname as member, facs . name as facility
	from 
		cd . members mems
		inner join cd . bookings bks
			on mems . memid = bks . memid
		inner join cd . facilities facs
			on bks . facid = facs . facid
	where
		bks . facid in ( 0 , 1 )
order by member          

Diese Übung ist weitgehend eine komplexere Anwendung dessen, was Sie in früheren Fragen gelernt haben. Es ist auch das erste Mal, dass wir mehr als einen Join verwendet haben, was für einige ein wenig verwirrend sein kann. Denken Sie beim Lesen von Join -Ausdrücken daran, dass ein Join effektiv eine Funktion ist, die zwei Tabellen aufnimmt, eine mit der linken Tabelle und die andere die rechte. Dies ist leicht zu visualisieren mit nur einem Anschluss an der Abfrage, aber ein wenig verwirrender mit zwei.

Unsere zweite INNER JOIN in dieser Abfrage hat eine rechte Seite von CD.Facities. Das ist leicht genug zu verstehen. Die linke Seite ist jedoch die Tabelle, die durch den Beitritt zu CD.Members zu CD.Bookings zurückgegeben wird. Es ist wichtig, dies zu betonen: Das relationale Modell dreht sich alles um Tabellen. Die Ausgabe eines Join ist eine weitere Tabelle. Die Ausgabe einer Abfrage ist eine Tabelle. Single Columned Listen sind Tabellen. Sobald Sie das erfasst haben, haben Sie die grundlegende Schönheit des Modells erfasst.

Als letzte Notiz stellen wir hier eine neue Sache vor: das || Der Bediener wird verwendet, um Saiten zu verkettet.

Erstellen Sie eine Liste kostspieliger Buchungen

Wie können Sie am Tag 2012-09-14 eine Liste von Buchungen erstellen, die das Mitglied (oder Gast) mehr als 30 US-Dollar kosten? Denken Sie daran, dass die Gäste unterschiedliche Kosten für Mitglieder haben (die aufgelisteten Kosten sind pro halbstündiger 'Slot'), und der Gastbenutzer ist immer id 0. In Ihren Ausgang den Namen der Einrichtung, den Namen des als einzelnen Spalte formierten Mitglieds und die Kosten einbeziehen. Bestellen Sie durch absteigende Kosten und verwenden Sie keine Unterabfragen.

Erwartete Ergebnisse:

Mitglied	Einrichtung	kosten
Gast	Massageraum 2	320
Gast	Massageraum 1	160
Gast	Massageraum 1	160
Gast	Massageraum 1	160
Gast	Tennisplatz 2	150
Jemima Farrell	Massageraum 1	140
Gast	Tennisplatz 1	75
Gast	Tennisplatz 2	75
Gast	Tennisplatz 1	75
Matthew Genting	Massageraum 1	70
Florence Bader	Massageraum 2	70
Gast	Kürbisgericht	70.0
Jemima Farrell	Massageraum 1	70
Überlegen Sie Stibbons	Massageraum 1	70
Burton Tracy	Massageraum 1	70
Jack Smith	Massageraum 1	70
Gast	Kürbisgericht	35.0
Gast	Kürbisgericht	35.0

Antwort:

 select mems . firstname || ' ' || mems . surname as member, 
	facs . name as facility, 
	case 
		when mems . memid = 0 then
			bks . slots * facs . guestcost
		else
			bks . slots * facs . membercost
	end as cost
        from
                cd . members mems                
                inner join cd . bookings bks
                        on mems . memid = bks . memid
                inner join cd . facilities facs
                        on bks . facid = facs . facid
        where
		bks . starttime >= ' 2012-09-14 ' and 
		bks . starttime < ' 2012-09-15 ' and (
			( mems . memid = 0 and bks . slots * facs . guestcost > 30 ) or
			( mems . memid != 0 and bks . slots * facs . membercost > 30 )
		)
order by cost desc ;          

Dies ist ein bisschen kompliziert! Obwohl es eine komplexere Logik als zuvor verwendet hat, gibt es keine schreckliche Bemerkung. Die WHERE beschränkt unsere Ausgabe auf ausreichend kostspielige Zeilen am 09.09.2012 und erinnert sich daran, zwischen Gästen und anderen zu unterscheiden. Wir verwenden dann eine CASE in der Spaltenauswahl, um die richtigen Kosten für das Mitglied oder Gast auszugeben.

Erstellen Sie eine Liste aller Mitglieder zusammen mit ihrem Empfehlend

Wie können Sie eine Liste aller Mitglieder, einschließlich der Person ausgeben, die sie empfohlen hat (falls vorhanden), ohne Verknüpfungen zu verwenden? Stellen Sie sicher, dass in der Liste keine Duplikate vorhanden sind und dass jede Erstname + -nurname -Paarung als Spalte formatiert und geordnet ist.

Erwartete Ergebnisse:

Mitglied	Empfohlen
Anna Mackenzie	Darren Smith
Anne Baker	Überlegen Sie Stibbons
Burton Tracy
Charles Owen	Darren Smith
Darren Smith
David Farrell
David Jones	Janice Joplette
David Pinker	Jemima Farrell
Douglas Jones	David Jones
Erica Crumpet	Tracy Smith
Florence Bader	Überlegen Sie Stibbons
Gast
Gerald Butters	Darren Smith
Henrietta Rumney	Matthew Genting
Henry Worthington-Smyth	Tracy Smith
Hyacinth Tupperware
Jack Smith	Darren Smith
Janice Joplette	Darren Smith
Jemima Farrell
Joan Coplin	Timothy Baker
John Hunt	Millicent -Zuständigkeitsbereich
Matthew Genting	Gerald Butters
Millicent -Zuständigkeitsbereich	Tracy Smith
Nancy Dare	Janice Joplette
Überlegen Sie Stibbons	Burton Tracy
Ramnaresh Sarwin	Florence Bader
Tim Boothe	Tim Rownam
Tim Rownam
Timothy Baker	Jemima Farrell
Tracy Smith

Antwort:

 select distinct mems . firstname || ' ' ||  mems . surname as member,
	( select recs . firstname || ' ' || recs . surname as recommender 
		from cd . members recs 
		where recs . memid = mems . recommendedby
	)
	from 
		cd . members mems
order by member;          

This exercise marks the introduction of subqueries. Subqueries are, as the name implies, queries within a query. They're commonly used with aggregates, to answer questions like 'get me all the details of the member who has spent the most hours on Tennis Court 1'.

In this case, we're simply using the subquery to emulate an outer join. For every value of member, the subquery is run once to find the name of the individual who recommended them (if any). A subquery that uses information from the outer query in this way (and thus has to be run for each row in the result set) is known as a correlated subquery .

Produce a list of costly bookings, using a subquery

The Produce a list of costly bookings exercise contained some messy logic: we had to calculate the booking cost in both the WHERE clause and the CASE statement. Try to simplify this calculation using subqueries. For reference, the question was:

How can you produce a list of bookings on the day of 2012-09-14 which will cost the member (or guest) more than $30? Remember that guests have different costs to members (the listed costs are per half-hour 'slot'), and the guest user is always ID 0. Include in your output the name of the facility, the name of the member formatted as a single column, and the cost. Order by descending cost.

Expected results:

Mitglied	Einrichtung	kosten
GUEST GUEST	Massage Room 2	320
GUEST GUEST	Massage Room 1	160
GUEST GUEST	Massage Room 1	160
GUEST GUEST	Massage Room 1	160
GUEST GUEST	Tennis Court 2	150
Jemima Farrell	Massage Room 1	140
GUEST GUEST	Tennis Court 1	75
GUEST GUEST	Tennis Court 2	75
GUEST GUEST	Tennis Court 1	75
Matthew Genting	Massage Room 1	70
Florence Bader	Massage Room 2	70
GUEST GUEST	Squash Court	70.0
Jemima Farrell	Massage Room 1	70
Ponder Stibbons	Massage Room 1	70
Burton Tracy	Massage Room 1	70
Jack Smith	Massage Room 1	70
GUEST GUEST	Squash Court	35.0
GUEST GUEST	Squash Court	35.0

Antwort:

 select member, facility, cost from (
	select 
		mems . firstname || ' ' || mems . surname as member,
		facs . name as facility,
		case
			when mems . memid = 0 then
				bks . slots * facs . guestcost
			else
				bks . slots * facs . membercost
		end as cost
		from
			cd . members mems
			inner join cd . bookings bks
				on mems . memid = bks . memid
			inner join cd . facilities facs
				on bks . facid = facs . facid
		where
			bks . starttime >= ' 2012-09-14 ' and
			bks . starttime < ' 2012-09-15 '
	) as bookings
	where cost > 30
order by cost desc ;          

This answer provides a mild simplification to the previous iteration: in the no-subquery version, we had to calculate the member or guest's cost in both the WHERE clause and the CASE statement. In our new version, we produce an inline query that calculates the total booking cost for us, allowing the outer query to simply select the bookings it's looking for. For reference, you may also see subqueries in the FROM clause referred to as inline views .

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Modifying Data

Querying data is all well and good, but at some point you're probably going to want to put data into your database! This section deals with inserting, updating, and deleting information. Operations that alter your data like this are collectively known as Data Manipulation Language, or DML.

In previous sections, we returned to you the results of the query you've performed. Since modifications like the ones we're making in this section don't return any query results, we instead show you the updated content of the table you're supposed to be working on. You can compare this with the table shown in 'Expected Results' to see how you've done.

If you struggle with these questions, I strongly recommend Learning SQL, by Alan Beaulieu.

Insert some data into a table

The club is adding a new facility - a spa. We need to add it into the facilities table. Use the following values:

• facid: 9, Name: 'Spa', membercost: 20, guestcost: 30, initialoutlay: 100000, monthlymaintenance: 800.

Expected results:

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	5	25	10000	200
1	Tennis Court 2	5	25	8000	200
2	Badminton Court	0	15.5	4000	50
3	Tischtennis	0	5	320	10
4	Massage Room 1	35	80	4000	3000
5	Massage Room 2	35	80	4000	3000
6	Squash Court	3.5	17.5	5000	80
7	Snooker Table	0	5	450	15
8	Pool Table	0	5	400	15
9	Spa	20	30	100000	800

Antwort:

 insert into cd . facilities
    (facid, name, membercost, guestcost, initialoutlay, monthlymaintenance)
    values ( 9 , ' Spa ' , 20 , 30 , 100000 , 800 );          

INSERT INTO ... VALUES is the simplest way to insert data into a table. There's not a whole lot to discuss here: VALUES is used to construct a row of data, which the INSERT statement inserts into the table. It's a simple as that.

You can see that there's two sections in parentheses. The first is part of the INSERT statement, and specifies the columns that we're providing data for. The second is part of VALUES , and specifies the actual data we want to insert into each column.

If we're inserting data into every column of the table, as in this example, explicitly specifying the column names is optional. As long as you fill in data for all columns of the table, in the order they were defined when you created the table, you can do something like the following:

 insert into cd . facilities values ( 9 , ' Spa ' , 20 , 30 , 100000 , 800 );

Generally speaking, for SQL that's going to be reused I tend to prefer being explicit and specifying the column names.

Insert multiple rows of data into a table

In the previous exercise, you learned how to add a facility. Now you're going to add multiple facilities in one command. Use the following values:

• facid: 9, Name: 'Spa', membercost: 20, guestcost: 30, initialoutlay: 100000, monthlymaintenance: 800.
• facid: 10, Name: 'Squash Court 2', membercost: 3.5, guestcost: 17.5, initialoutlay: 5000, monthlymaintenance: 80.

Expected results:

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	5	25	10000	200
1	Tennis Court 2	5	25	8000	200
2	Badminton Court	0	15.5	4000	50
3	Tischtennis	0	5	320	10
4	Massage Room 1	35	80	4000	3000
5	Massage Room 2	35	80	4000	3000
6	Squash Court	3.5	17.5	5000	80
7	Snooker Table	0	5	450	15
8	Pool Table	0	5	400	15
9	Spa	20	30	100000	800
10	Squash Court 2	3.5	17.5	5000	80

Antwort:

 insert into cd . facilities
    (facid, name, membercost, guestcost, initialoutlay, monthlymaintenance)
    values
        ( 9 , ' Spa ' , 20 , 30 , 100000 , 800 ),
        ( 10 , ' Squash Court 2 ' , 3 . 5 , 17 . 5 , 5000 , 80 );          

VALUES can be used to generate more than one row to insert into a table, as seen in this example. Hopefully it's clear what's going on here: the output of VALUES is a table, and that table is copied into cd.facilities, the table specified in the INSERT command.

While you'll most commonly see VALUES when inserting data, Postgres allows you to use VALUES wherever you might use a SELECT . This makes sense: the output of both commands is a table, it's just that VALUES is a bit more ergonomic when working with constant data.

Similarly, it's possible to use SELECT wherever you see a VALUES . This means that you can INSERT the results of a SELECT . Zum Beispiel:

 insert into cd . facilities
    (facid, name, membercost, guestcost, initialoutlay, monthlymaintenance)
    SELECT 9 , ' Spa ' , 20 , 30 , 100000 , 800
    UNION ALL
        SELECT 10 , ' Squash Court 2 ' , 3 . 5 , 17 . 5 , 5000 , 80 ;

In later exercises you'll see us using INSERT ... SELECT to generate data to insert based on the information already in the database.

Insert calculated data into a table

Let's try adding the spa to the facilities table again. This time, though, we want to automatically generate the value for the next facid, rather than specifying it as a constant. Use the following values for everything else:

• Name: 'Spa', membercost: 20, guestcost: 30, initialoutlay: 100000, monthlymaintenance: 800.

Expected results:

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	5	25	10000	200
1	Tennis Court 2	5	25	8000	200
2	Badminton Court	0	15.5	4000	50
3	Tischtennis	0	5	320	10
4	Massage Room 1	35	80	4000	3000
5	Massage Room 2	35	80	4000	3000
6	Squash Court	3.5	17.5	5000	80
7	Snooker Table	0	5	450	15
8	Pool Table	0	5	400	15
9	Spa	20	30	100000	800

Antwort:

 insert into cd . facilities
    (facid, name, membercost, guestcost, initialoutlay, monthlymaintenance)
    select ( select max (facid) from cd . facilities ) + 1 , ' Spa ' , 20 , 30 , 100000 , 800 ;          

In the previous exercises we used VALUES to insert constant data into the facilities table. Here, though, we have a new requirement: a dynamically generated ID. This gives us a real quality of life improvement, as we don't have to manually work out what the current largest ID is: the SQL command does it for us.

Since the VALUES clause is only used to supply constant data, we need to replace it with a query instead. The SELECT statement is fairly simple: there's an inner subquery that works out the next facid based on the largest current id, and the rest is just constant data. The output of the statement is a row that we insert into the facilities table.

While this works fine in our simple example, it's not how you would generally implement an incrementing ID in the real world. Postgres provides SERIAL types that are auto-filled with the next ID when you insert a row. As well as saving us effort, these types are also safer: unlike the answer given in this exercise, there's no need to worry about concurrent operations generating the same ID.

Update some existing data

We made a mistake when entering the data for the second tennis court. The initial outlay was 10000 rather than 8000: you need to alter the data to fix the error.

Expected results:

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	5	25	10000	200
1	Tennis Court 2	5	25	10000	200
2	Badminton Court	0	15.5	4000	50
3	Tischtennis	0	5	320	10
4	Massage Room 1	35	80	4000	3000
5	Massage Room 2	35	80	4000	3000
6	Squash Court	3.5	17.5	5000	80
7	Snooker Table	0	5	450	15
8	Pool Table	0	5	400	15

Antwort:

 update cd . facilities
    set initialoutlay = 10000
    where facid = 1 ;          

The UPDATE statement is used to alter existing data. If you're familiar with SELECT queries, it's pretty easy to read: the WHERE clause works in exactly the same fashion, allowing us to filter the set of rows we want to work with. These rows are then modified according to the specifications of the SET clause: in this case, setting the initial outlay.

The WHERE clause is extremely important. It's easy to get it wrong or even omit it, with disastrous results. Consider the following command:

 update cd . facilities
    set initialoutlay = 10000 ;

There's no WHERE clause to filter for the rows we're interested in. The result of this is that the update runs on every row in the table! This is rarely what we want to happen.

Update multiple rows and columns at the same time

We want to increase the price of the tennis courts for both members and guests. Update the costs to be 6 for members, and 30 for guests.

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	6	30	10000	200
1	Tennis Court 2	6	30	8000	200
2	Badminton Court	0	15.5	4000	50
3	Tischtennis	0	5	320	10
4	Massage Room 1	35	80	4000	3000
5	Massage Room 2	35	80	4000	3000
6	Squash Court	3.5	17.5	5000	80
7	Snooker Table	0	5	450	15
8	Pool Table	0	5	400	15

Antwort:

 update cd . facilities
    set
        membercost = 6 ,
        guestcost = 30
    where facid in ( 0 , 1 );          

The SET clause accepts a comma separated list of values that you want to update.

Update a row based on the contents of another row

We want to alter the price of the second tennis court so that it costs 10% more than the first one. Try to do this without using constant values for the prices, so that we can reuse the statement if we want to.

Expected results:

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	5	25	10000	200
1	Tennis Court 2	5.5	27.5	8000	200
2	Badminton Court	0	15.5	4000	50
3	Tischtennis	0	5	320	10
4	Massage Room 1	35	80	4000	3000
5	Massage Room 2	35	80	4000	3000
6	Squash Court	3.5	17.5	5000	80
7	Snooker Table	0	5	450	15
8	Pool Table	0	5	400	15

Antwort:

 update cd . facilities facs
    set
        membercost = ( select membercost * 1 . 1 from cd . facilities where facid = 0 ),
        guestcost = ( select guestcost * 1 . 1 from cd . facilities where facid = 0 )
    where facs . facid = 1 ;          

Updating columns based on calculated data is not too intrinsically difficult: we can do so pretty easily using subqueries. You can see this approach in our selected answer.

As the number of columns we want to update increases, standard SQL can start to get pretty awkward: you don't want to be specifying a separate subquery for each of 15 different column updates. Postgres provides a nonstandard extension to SQL called UPDATE...FROM that addresses this: it allows you to supply a FROM clause to generate values for use in the SET clause. Example below:

 update cd . facilities facs
    set
        membercost = facs2 . membercost * 1 . 1 ,
        guestcost = facs2 . guestcost * 1 . 1
    from ( select * from cd . facilities where facid = 0 ) facs2
    where facs . facid = 1 ;

Delete all bookings

As part of a clearout of our database, we want to delete all bookings from the cd.bookings table. How can we accomplish this?

Expected results:

bookid	facid	memid	starttime	slots

Antwort:

 delete from cd . bookings ;          

The DELETE statement does what it says on the tin: deletes rows from the table. Here, we show the command in its simplest form, with no qualifiers. In this case, it deletes everything from the table. Obviously, you should be careful with your deletes and make sure they're always limited - we'll see how to do that in the next exercise.

An alternative to unqualified DELETE s is the following:

truncate cd . bookings ;

TRUNCATE also deletes everything in the table, but does so using a quicker underlying mechanism. It's not perfectly safe in all circumstances, though, so use judiciously. When in doubt, use DELETE .

Delete a member from the cd.members table

We want to remove member 37, who has never made a booking, from our database. How can we achieve that?

Expected results:

memid	Nachname	firstname	Adresse	PLZ	Telefon	Empfohlen von	joindate
0	GAST	GAST	GAST	0	[phone]		2012-07-01 00:00:00
1	Schmied	Darren	8 Bloomsbury Close, Boston	4321	555-555-5555		2012-07-02 12:02:05
2	Schmied	Tracy	8 Bloomsbury Close, New York	4321	555-555-5555		2012-07-02 12:08:23
3	Rownam	Tim	23 Highway Way, Boston	23423	[phone]		2012-07-03 09:32:15
4	Joplette	Janice	20 Crossing Road, New York	234	[phone]	1	2012-07-03 10:25:05
5	Butters	Gerald	1065 Huntingdon Avenue, Boston	56754	[phone]	1	2012-07-09 10:44:09
6	Tracy	Burton	3 Tunisia Drive, Boston	45678	[phone]		2012-07-15 08:52:55
7	Wagen	Nancy	6 Hunting Lodge Way, Boston	10383	[phone]	4	2012-07-25 08:59:12
8	Boothe	Tim	3 Bloomsbury Close, Reading, 00234	234	[phone]	3	2012-07-25 16:02:35
9	Stibbons	Ponder	5 Dragons Way, Winchester	87630	[phone]	6	2012-07-25 17:09:05
10	Owen	Charles	52 Cheshire Grove, Winchester, 28563	28563	[phone]	1	2012-08-03 19:42:37
11	Jones	David	976 Gnats Close, Reading	33862	[phone]	4	2012-08-06 16:32:55
12	Bäcker	Anne	55 Powdery Street, Boston	80743	844-076-5141	9	2012-08-10 14:23:22
13	Farrell	Jemima	103 Firth Avenue, North Reading	57392	[phone]		2012-08-10 14:28:01
14	Schmied	Jack	252 Binkington Way, Boston	69302	[phone]	1	2012-08-10 16:22:05
15	Bader	Florenz	264 Ursula Drive, Westford	84923	[phone]	9	2012-08-10 17:52:03
16	Bäcker	Timothy	329 James Street, Reading	58393	833-941-0824	13	2012-08-15 10:34:25
17	Pinker	David	5 Impreza Road, Boston	65332	811 409-6734	13	2012-08-16 11:32:47
20	Genting	Matthew	4 Nunnington Place, Wingfield, Boston	52365	[phone]	5	2012-08-19 14:55:55
21	Mackenzie	Anna	64 Perkington Lane, Reading	64577	[phone]	1	2012-08-26 09:32:05
22	Coplin	Joan	85 Bard Street, Bloomington, Boston	43533	[phone]	16	2012-08-29 08:32:41
24	Sarwin	Ramnaresh	12 Bullington Lane, Boston	65464	[phone]	15	2012-09-01 08:44:42
26	Jones	Douglas	976 Gnats Close, Reading	11986	844 536-8036	11	2012-09-02 18:43:05
27	Rumney	Henrietta	3 Burkington Plaza, Boston	78533	[phone]	20	2012-09-05 08:42:35
28	Farrell	David	437 Granite Farm Road, Westford	43532	[phone]		2012-09-15 08:22:05
29	Worthington-Smyth	Henry	55 Jagbi Way, North Reading	97676	[phone]	2	2012-09-17 12:27:15
30	Aufgabe	Millicent	641 Drudgery Close, Burnington, Boston	34232	[phone]	2	2012-09-18 19:04:01
33	Tupperware	Hyazinthe	33 Cheerful Plaza, Drake Road, Westford	68666	[phone]		2012-09-18 19:32:05
35	Jagd	John	5 Bullington Lane, Boston	54333	[phone]	30	2012-09-19 11:32:45
36	Crumpet	Erica	Crimson Road, North Reading	75655	[phone]	2	2012-09-22 08:36:38

Antwort:

 delete from cd . members where memid = 37 ;          

This exercise is a small increment on our previous one. Instead of deleting all bookings, this time we want to be a bit more targeted, and delete a single member that has never made a booking. To do this, we simply have to add a WHERE clause to our command, specifying the member we want to delete. You can see the parallels with SELECT and UPDATE statements here.

There's one interesting wrinkle here. Try this command out, but substituting in member id 0 instead. This member has made many bookings, and you'll find that the delete fails with an error about a foreign key constraint violation. This is an important concept in relational databases, so let's explore a little further.

Foreign keys are a mechanism for defining relationships between columns of different tables. In our case we use them to specify that the memid column of the bookings table is related to the memid column of the members table. The relationship (or 'constraint') specifies that for a given booking, the member specified in the booking must exist in the members table. It's useful to have this guarantee enforced by the database: it means that code using the database can rely on the presence of the member. It's hard (even impossible) to enforce this at higher levels: concurrent operations can interfere and leave your database in a broken state.

PostgreSQL supports various different kinds of constraints that allow you to enforce structure upon your data. For more information on constraints, check out the PostgreSQL documentation on foreign keys

Delete based on a subquery

In our previous exercises, we deleted a specific member who had never made a booking. How can we make that more general, to delete all members who have never made a booking?

Expected results:

memid	Nachname	firstname	Adresse	PLZ	Telefon	Empfohlen von	joindate
0	GAST	GAST	GAST	0	[phone]		2012-07-01 00:00:00
1	Schmied	Darren	8 Bloomsbury Close, Boston	4321	555-555-5555		2012-07-02 12:02:05
2	Schmied	Tracy	8 Bloomsbury Close, New York	4321	555-555-5555		2012-07-02 12:08:23
3	Rownam	Tim	23 Highway Way, Boston	23423	[phone]		2012-07-03 09:32:15
4	Joplette	Janice	20 Crossing Road, New York	234	[phone]	1	2012-07-03 10:25:05
5	Butters	Gerald	1065 Huntingdon Avenue, Boston	56754	[phone]	1	2012-07-09 10:44:09
6	Tracy	Burton	3 Tunisia Drive, Boston	45678	[phone]		2012-07-15 08:52:55
7	Wagen	Nancy	6 Hunting Lodge Way, Boston	10383	[phone]	4	2012-07-25 08:59:12
8	Boothe	Tim	3 Bloomsbury Close, Reading, 00234	234	[phone]	3	2012-07-25 16:02:35
9	Stibbons	Ponder	5 Dragons Way, Winchester	87630	[phone]	6	2012-07-25 17:09:05
10	Owen	Charles	52 Cheshire Grove, Winchester, 28563	28563	[phone]	1	2012-08-03 19:42:37
11	Jones	David	976 Gnats Close, Reading	33862	[phone]	4	2012-08-06 16:32:55
12	Bäcker	Anne	55 Powdery Street, Boston	80743	844-076-5141	9	2012-08-10 14:23:22
13	Farrell	Jemima	103 Firth Avenue, North Reading	57392	[phone]		2012-08-10 14:28:01
14	Schmied	Jack	252 Binkington Way, Boston	69302	[phone]	1	2012-08-10 16:22:05
15	Bader	Florenz	264 Ursula Drive, Westford	84923	[phone]	9	2012-08-10 17:52:03
16	Bäcker	Timothy	329 James Street, Reading	58393	833-941-0824	13	2012-08-15 10:34:25
17	Pinker	David	5 Impreza Road, Boston	65332	811 409-6734	13	2012-08-16 11:32:47
20	Genting	Matthew	4 Nunnington Place, Wingfield, Boston	52365	[phone]	5	2012-08-19 14:55:55
21	Mackenzie	Anna	64 Perkington Lane, Reading	64577	[phone]	1	2012-08-26 09:32:05
22	Coplin	Joan	85 Bard Street, Bloomington, Boston	43533	[phone]	16	2012-08-29 08:32:41
24	Sarwin	Ramnaresh	12 Bullington Lane, Boston	65464	[phone]	15	2012-09-01 08:44:42
26	Jones	Douglas	976 Gnats Close, Reading	11986	844 536-8036	11	2012-09-02 18:43:05
27	Rumney	Henrietta	3 Burkington Plaza, Boston	78533	[phone]	20	2012-09-05 08:42:35
28	Farrell	David	437 Granite Farm Road, Westford	43532	[phone]		2012-09-15 08:22:05
29	Worthington-Smyth	Henry	55 Jagbi Way, North Reading	97676	[phone]	2	2012-09-17 12:27:15
30	Aufgabe	Millicent	641 Drudgery Close, Burnington, Boston	34232	[phone]	2	2012-09-18 19:04:01
33	Tupperware	Hyazinthe	33 Cheerful Plaza, Drake Road, Westford	68666	[phone]		2012-09-18 19:32:05
35	Jagd	John	5 Bullington Lane, Boston	54333	[phone]	30	2012-09-19 11:32:45
36	Crumpet	Erica	Crimson Road, North Reading	75655	[phone]	2	2012-09-22 08:36:38

Antwort:

 delete from cd . members where memid not in ( select memid from cd . bookings );          

We can use subqueries to determine whether a row should be deleted or not. There's a couple of standard ways to do this. In our featured answer, the subquery produces a list of all the different member ids in the cd.bookings table. If a row in the table isn't in the list generated by the subquery, it gets deleted.

An alternative is to use a correlated subquery . Where our previous example runs a large subquery once, the correlated approach instead specifies a smaller subqueryto run against every row.

 delete from cd . members mems where not exists ( select 1 from cd . bookings where memid = mems . memid );

The two different forms can have different performance characteristics. Under the hood, your database engine is free to transform your query to execute it in a correlated or uncorrelated fashion, though, so things can be a little hard to predict.

---

Aggregation

Aggregation is one of those capabilities that really make you appreciate the power of relational database systems. It allows you to move beyond merely persisting your data, into the realm of asking truly interesting questions that can be used to inform decision making. This category covers aggregation at length, making use of standard grouping as well as more recent window functions.

If you struggle with these questions, I strongly recommend Learning SQL, by Alan Beaulieu and SQL Cookbook by Anthony Molinaro. In fact, get the latter anyway - it'll take you beyond anything you find on this site, and on multiple different database systems to boot.

Count the number of facilities

For our first foray into aggregates, we're going to stick to something simple. We want to know how many facilities exist - simply produce a total count.

Expected results:

zählen
9

Antwort:

 select count ( * ) from cd . facilities ;          

Aggregation starts out pretty simply! The SQL above selects everything from our facilities table, and then counts the number of rows in the result set. The count function has a variety of uses:

• COUNT(*) simply returns the number of rows
• COUNT(address) counts the number of non-null addresses in the result set.
• Finally, COUNT(DISTINCT address) counts the number of different addresses in the facilities table.

The basic idea of an aggregate function is that it takes in a column of data, performs some function upon it, and outputs a scalar (single) value. There are a bunch more aggregation functions, including MAX , MIN , SUM , and AVG . These all do pretty much what you'd expect from their names :-).

One aspect of aggregate functions that people often find confusing is in queries like the below:

 select facid, count ( * ) from cd . facilities

Try it out, and you'll find that it doesn't work. This is because count(*) wants to collapse the facilities table into a single value - unfortunately, it can't do that, because there's a lot of different facids in cd.facilities - Postgres doesn't know which facid to pair the count with.

Instead, if you wanted a query that returns all the facids along with a count on each row, you can break the aggregation out into a subquery as below:

 select facid, 
	( select count ( * ) from cd . facilities )
	from cd . facilities

When we have a subquery that returns a scalar value like this, Postgres knows to simply repeat the value for every row in cd.facilities.

Count the number of expensive facilities

Produce a count of the number of facilities that have a cost to guests of 10 or more.

zählen
6

Antwort:

 select count ( * ) from cd . facilities where guestcost >= 10 ;          

This one is only a simple modification to the previous question: we need to weed out the inexpensive facilities. This is easy to do using a WHERE clause. Our aggregation can now only see the expensive facilities.

Count the number of recommendations each member makes

Produce a count of the number of recommendations each member has made. Order by member ID.

Expected results:

Empfohlen von	zählen
1	5
2	3
3	1
4	2
5	1
6	1
9	2
11	1
13	2
15	1
16	1
20	1
30	1

Antwort:

 select recommendedby, count ( * ) 
	from cd . members
	where recommendedby is not null
	group by recommendedby
order by recommendedby;          

Previously, we've seen that aggregation functions are applied to a column of values, and convert them into an aggregated scalar value. This is useful, but we often find that we don't want just a single aggregated result: for example, instead of knowing the total amount of money the club has made this month, I might want to know how much money each different facility has made, or which times of day were most lucrative.

In order to support this kind of behaviour, SQL has the GROUP BY construct. What this does is batch the data together into groups, and run the aggregation function separately for each group. When you specify a GROUP BY , the database produces an aggregated value for each distinct value in the supplied columns. In this case, we're saying 'for each distinct value of recommendedby, get me the number of times that value appears'.

List the total slots booked per facility

Produce a list of the total number of slots booked per facility. For now, just produce an output table consisting of facility id and slots, sorted by facility id.

Expected results:

facid	Total Slots
0	1320
1	1278
2	1209
3	830
4	1404
5	228
6	1104
7	908
8	911

Antwort:

 select facid, sum (slots) as " Total Slots "
	from cd . bookings
	group by facid
order by facid;          

Other than the fact that we've introduced the SUM aggregate function, there's not a great deal to say about this exercise. For each distinct facility id, the SUM function adds together everything in the slots column.

List the total slots booked per facility in a given month

Produce a list of the total number of slots booked per facility in the month of September 2012. Produce an output table consisting of facility id and slots, sorted by the number of slots.

Expected results:

facid	Total Slots
5	122
3	422
7	426
8	471
6	540
2	570
1	588
0	591
4	648

Antwort:

 select facid, sum (slots) as " Total Slots "
	from cd . bookings
	where
		starttime >= ' 2012-09-01 '
		and starttime < ' 2012-10-01 '
	group by facid
order by sum (slots);          

This is only a minor alteration of our previous example. Remember that aggregation happens after the WHERE clause is evaluated: we thus use the WHERE to restrict the data we aggregate over, and our aggregation only sees data from a single month.

List the total slots booked per facility per month

Produce a list of the total number of slots booked per facility per month in the year of 2012. Produce an output table consisting of facility id and slots, sorted by the id and month.

Expected results:

facid	Monat	Total Slots
0	7	270
0	8	459
0	9	591
1	7	207
1	8	483
1	9	588
2	7	180
2	8	459
2	9	570
3	7	104
3	8	304
3	9	422
4	7	264
4	8	492
4	9	648
5	7	24
5	8	82
5	9	122
6	7	164
6	8	400
6	9	540
7	7	156
7	8	326
7	9	426
8	7	117
8	8	322
8	9	471

Antwort:

 select facid, extract(month from starttime) as month, sum (slots) as " Total Slots "
	from cd . bookings
	where
		starttime >= ' 2012-01-01 '
		and starttime < ' 2013-01-01 '
	group by facid, month
order by facid, month;          

The main piece of new functionality in this question is the EXTRACT function. EXTRACT allows you to get individual components of a timestamp, like day, month, year, etc. We group by the output of this function to provide per-month values. An alternative, if we needed to distinguish between the same month in different years, is to make use of the DATE_TRUNC function, which truncates a date to a given granularity.

It's also worth noting that this is the first time we've truly made use of the ability to group by more than one column.

Find the count of members who have made at least one booking

Find the total number of members who have made at least one booking.

Expected results:

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30

Antwort:

 select count (distinct memid) from cd . bookings          

Your first instinct may be to go for a subquery here. Something like the below:

 select count ( * ) from 
	( select distinct memid from cd . bookings ) as mems

This does work perfectly well, but we can simplify a touch with the help of a little extra knowledge in the form of COUNT DISTINCT . This does what you might expect, counting the distinct values in the passed column.

List facilities with more than 1000 slots booked

Produce a list of facilities with more than 1000 slots booked. Produce an output table consisting of facility id and hours, sorted by facility id.

Expected results:

facid	Total Slots
0	1320
1	1278
2	1209
4	1404
6	1104

Antwort:

 select facid, sum (slots) as " Total Slots "
        from cd . bookings
        group by facid
        having sum (slots) > 1000
        order by facid          

It turns out that there's actually an SQL keyword designed to help with the filtering of output from aggregate functions. This keyword is HAVING .

The behaviour of HAVING is easily confused with that of WHERE . The best way to think about it is that in the context of a query with an aggregate function, WHERE is used to filter what data gets input into the aggregate function, while HAVING is used to filter the data once it is output from the function. Try experimenting to explore this difference!

Find the total revenue of each facility

Produce a list of facilities along with their total revenue. The output table should consist of facility name and revenue, sorted by revenue. Remember that there's a different cost for guests and members!

Expected results:

Name	Einnahmen
Tischtennis	180
Snooker Table	240
Pool Table	270
Badminton Court	1906.5
Squash Court	13468.0
Tennis Court 1	13860
Tennis Court 2	14310
Massage Room 2	15810
Massage Room 1	72540

Antwort:

 select facs . name , sum (slots * case
			when memid = 0 then facs . guestcost
			else facs . membercost
		end) as revenue
	from cd . bookings bks
	inner join cd . facilities facs
		on bks . facid = facs . facid
	group by facs . name
order by revenue;          

The only real complexity in this query is that guests (member ID 0) have a different cost to everyone else. We use a case statement to produce the cost for each session, and then sum each of those sessions, grouped by facility.

Find facilities with a total revenue less than 1000

Produce a list of facilities with a total revenue less than 1000. Produce an output table consisting of facility name and revenue, sorted by revenue. Remember that there's a different cost for guests and members!

Expected results:

Name	Einnahmen
Tischtennis	180
Snooker Table	240
Pool Table	270

Antwort:

 select name, revenue from (
	select facs . name , sum (case 
				when memid = 0 then slots * facs . guestcost
				else slots * membercost
			end) as revenue
		from cd . bookings bks
		inner join cd . facilities facs
			on bks . facid = facs . facid
		group by facs . name
	) as agg where revenue < 1000
order by revenue;          

You may well have tried to use the HAVING keyword we introduced in an earlier exercise, producing something like below:

 select facs . name , sum (case 
		when memid = 0 then slots * facs . guestcost
		else slots * membercost
	end) as revenue
	from cd . bookings bks
	inner join cd . facilities facs
		on bks . facid = facs . facid
	group by facs . name
	having revenue < 1000
order by revenue;

Unfortunately, this doesn't work! You'll get an error along the lines of ERROR: column "revenue" does not exist . Postgres, unlike some other RDBMSs like SQL Server and MySQL, doesn't support putting column names in the HAVING clause. This means that for this query to work, you'd have to produce something like below:

 select facs . name , sum (case 
		when memid = 0 then slots * facs . guestcost
		else slots * membercost
	end) as revenue
	from cd . bookings bks
	inner join cd . facilities facs
		on bks . facid = facs . facid
	group by facs . name
	having sum (case 
		when memid = 0 then slots * facs . guestcost
		else slots * membercost
	end) < 1000
order by revenue;

Having to repeat significant calculation code like this is messy, so our anointed solution instead just wraps the main query body as a subquery, and selects from it using a WHERE clause. In general, I recommend using HAVING for simple queries, as it increases clarity. Otherwise, this subquery approach is often easier to use.

Output the facility id that has the highest number of slots booked

Output the facility id that has the highest number of slots booked. For bonus points, try a version without a LIMIT clause. This version will probably look messy!

Expected results:

facid	Total Slots
4	1404

Antwort:

 select facid, sum (slots) as " Total Slots "
	from cd . bookings
	group by facid
order by sum (slots) desc
LIMIT 1 ;          

Let's start off with what's arguably the simplest way to do this: produce a list of facility IDs and the total number of slots used, order by the total number of slots used, and pick only the top result.

It's worth realising, though, that this method has a significant weakness. In the event of a tie, we will still only get one result! To get all the relevant results, we might try using the MAX aggregate function, something like below:

 select facid, max (totalslots) from (
	select facid, sum (slots) as totalslots    
		from cd . bookings    
		group by facid
	) as sub group by facid

The intent of this query is to get the highest totalslots value and its associated facid(s). Unfortunately, this just won't work! In the event of multiple facids having the same number of slots booked, it would be ambiguous which facid should be paired up with the single (or scalar ) value coming out of the MAX function. This means that Postgres will tell you that facid ought to be in a GROUP BY section, which won't produce the results we're looking for.

Let's take a first stab at a working query:

 select facid, sum (slots) as totalslots
	from cd . bookings
	group by facid
	having sum (slots) = ( select max ( sum2 . totalslots ) from
		( select sum (slots) as totalslots
		from cd . bookings
		group by facid
		) as sum2);

The query produces a list of facility IDs and number of slots used, and then uses a HAVING clause that works out the maximum totalslots value. We're essentially saying: 'produce a list of facids and their number of slots booked, and filter out all the ones that doen't have a number of slots booked equal to the maximum.'

Useful as HAVING is, however, our query is pretty ugly. To improve on that, let's introduce another new concept: Common Table Expressions (CTEs). CTEs can be thought of as allowing you to define a database view inline in your query. It's really helpful in situations like this, where you're having to repeat yourself a lot.

CTEs are declared in the form WITH CTEName as (SQL-Expression) . You can see our query redefined to use a CTE below:

with sum as ( select facid, sum (slots) as totalslots
	from cd . bookings
	group by facid
)
select facid, totalslots 
	from sum
	where totalslots = ( select max (totalslots) from sum);

You can see that we've factored out our repeated selections from cd.bookings into a single CTE, and made the query a lot simpler to read in the process!

BUT WAIT. There's more. It's also possible to complete this problem using Window Functions. We'll leave these until later, but even better solutions to problems like these are available.

That's a lot of information for a single exercise. Don't worry too much if you don't get it all right now - we'll reuse these concepts in later exercises.

List the total slots booked per facility per month, Part 2

Produce a list of the total number of slots booked per facility per month in the year of 2012. In this version, include output rows containing totals for all months per facility, and a total for all months for all facilities. The output table should consist of facility id, month and slots, sorted by the id and month. When calculating the aggregated values for all months and all facids, return null values in the month and facid columns.

Expected results:

facid	Monat	slots
0	7	270
0	8	459
0	9	591
0		1320
1	7	207
1	8	483
1	9	588
1		1278
2	7	180
2	8	459
2	9	570
2		1209
3	7	104
3	8	304
3	9	422
3		830
4	7	264
4	8	492
4	9	648
4		1404
5	7	24
5	8	82
5	9	122
5		228
6	7	164
6	8	400
6	9	540
6		1104
7	7	156
7	8	326
7	9	426
7		908
8	7	117
8	8	322
8	9	471
8		910
		9191

Antwort:

 select facid, extract(month from starttime) as month, sum (slots) as slots
	from cd . bookings
	where
		starttime >= ' 2012-01-01 '
		and starttime < ' 2013-01-01 '
	group by rollup(facid, month)
order by facid, month;          

When we are doing data analysis, we sometimes want to perform multiple levels of aggregation to allow ourselves to 'zoom' in and out to different depths. In this case, we might be looking at each facility's overall usage, but then want to dive in to see how they've performed on a per-month basis. Using the SQL we know so far, it's quite cumbersome to produce a single query that does what we want - we effectively have to resort to concatenating multiple queries using UNION ALL :

 select facid, extract(month from starttime) as month, sum (slots) as slots
    from cd . bookings
    where
        starttime >= ' 2012-01-01 '
        and starttime < ' 2013-01-01 '
    group by facid, month
union all
select facid, null , sum (slots) as slots
    from cd . bookings
    where
        starttime >= ' 2012-01-01 '
        and starttime < ' 2013-01-01 '
    group by facid
union all
select null , null , sum (slots) as slots
    from cd . bookings
    where
        starttime >= ' 2012-01-01 '
        and starttime < ' 2013-01-01 '
order by facid, month;

As you can see, each subquery performs a different level of aggregation, and we just combine the results. We can clean this up a lot by factoring out commonalities using a CTE:

with bookings as (
	select facid, extract(month from starttime) as month, slots
	from cd . bookings
	where
		starttime >= ' 2012-01-01 '
		and starttime < ' 2013-01-01 '
)
select facid, month, sum (slots) from bookings group by facid, month
union all
select facid, null , sum (slots) from bookings group by facid
union all
select null , null , sum (slots) from bookings
order by facid, month;

This version is not excessively hard on the eyes, but it becomes cumbersome as the number of aggregation columns increases. Fortunately, PostgreSQL 9.5 introduced support for the ROLLUP operator, which we've used to simplify our accepted answer.

ROLLUP produces a hierarchy of aggregations in the order passed into it: for example, ROLLUP(facid, month) outputs aggregations on (facid, month), (facid), and (). If we wanted an aggregation of all facilities for a month (instead of all months for a facility) we'd have to reverse the order, using ROLLUP(month, facid) . Alternatively, if we instead want all possible permutations of the columns we pass in, we can use CUBE rather than ROLLUP . This will produce (facid, month), (month), (facid), and ().

ROLLUP and CUBE are special cases of GROUPING SETS . GROUPING SETS allow you to specify the exact aggregation permutations you want: you could, for example, ask for just (facid, month) and (facid), skipping the top-level aggregation.

List the total hours booked per named facility

Produce a list of the total number of hours booked per facility, remembering that a slot lasts half an hour. The output table should consist of the facility id, name, and hours booked, sorted by facility id. Try formatting the hours to two decimal places.

Expected results:

facid	Name	Total Hours
0	Tennis Court 1	660.00
1	Tennis Court 2	639.00
2	Badminton Court	604.50
3	Tischtennis	415.00
4	Massage Room 1	702.00
5	Massage Room 2	114.00
6	Squash Court	552.00
7	Snooker Table	454.00
8	Pool Table	455.50

Antwort:

 select facs . facid , facs . name ,
	trim (to_char( sum ( bks . slots ) / 2 . 0 , ' 9999999999999999D99 ' )) as " Total Hours "

	from cd . bookings bks
	inner join cd . facilities facs
		on facs . facid = bks . facid
	group by facs . facid , facs . name
order by facs . facid ;          

There's a few little pieces of interest in this question. Firstly, you can see that our aggregation works just fine when we join to another table on a 1:1 basis. Also note that we group by both facs.facid and facs.name . This is might seem odd: after all, since facid is the primary key of the facilities table, each facid has exactly one name, and grouping by both fields is the same as grouping by facid alone. In fact, you'll find that if you remove facs.name from the GROUP BY clause, the query works just fine: Postgres works out that this 1:1 mapping exists, and doesn't insist that we group by both columns.

Unfortunately, depending on which database system we use, validation might not be so smart, and may not realise that the mapping is strictly 1:1. That being the case, if there were multiple names for each facid and we hadn't grouped by name , the DBMS would have to choose between multiple (equally valid) choices for the name . Since this is invalid, the database system will insist that we group by both fields. In general, I recommend grouping by all columns you don't have an aggregate function on: this will ensure better cross-platform compatibility.

Next up is the division. Those of you familiar with MySQL may be aware that integer divisions are automatically cast to floats. Postgres is a little more traditional in this respect, and expects you to tell it if you want a floating point division. You can do that easily in this case by dividing by 2.0 rather than 2.

Finally, let's take a look at formatting. The TO_CHAR function converts values to character strings. It takes a formatting string, which we specify as (up to) lots of numbers before the decimal place, decimal place, and two numbers after the decimal place. The output of this function can be prepended with a space, which is why we include the outer TRIM function.

List each member's first booking after September 1st 2012

Produce a list of each member name, id, and their first booking after September 1st 2012. Order by member ID.

Expected results:

Nachname	firstname	memid	starttime
GAST	GAST	0	2012-09-01 08:00:00
Schmied	Darren	1	2012-09-01 09:00:00
Schmied	Tracy	2	2012-09-01 11:30:00
Rownam	Tim	3	2012-09-01 16:00:00
Joplette	Janice	4	2012-09-01 15:00:00
Butters	Gerald	5	2012-09-02 12:30:00
Tracy	Burton	6	2012-09-01 15:00:00
Wagen	Nancy	7	2012-09-01 12:30:00
Boothe	Tim	8	2012-09-01 08:30:00
Stibbons	Ponder	9	2012-09-01 11:00:00
Owen	Charles	10	2012-09-01 11:00:00
Jones	David	11	2012-09-01 09:30:00
Bäcker	Anne	12	2012-09-01 14:30:00
Farrell	Jemima	13	2012-09-01 09:30:00
Schmied	Jack	14	2012-09-01 11:00:00
Bader	Florenz	15	2012-09-01 10:30:00
Bäcker	Timothy	16	2012-09-01 15:00:00
Pinker	David	17	2012-09-01 08:30:00
Genting	Matthew	20	2012-09-01 18:00:00
Mackenzie	Anna	21	2012-09-01 08:30:00
Coplin	Joan	22	2012-09-02 11:30:00
Sarwin	Ramnaresh	24	2012-09-04 11:00:00
Jones	Douglas	26	2012-09-08 13:00:00
Rumney	Henrietta	27	2012-09-16 13:30:00
Farrell	David	28	2012-09-18 09:00:00
Worthington-Smyth	Henry	29	2012-09-19 09:30:00
Aufgabe	Millicent	30	2012-09-19 11:30:00
Tupperware	Hyazinthe	33	2012-09-20 08:00:00
Jagd	John	35	2012-09-23 14:00:00
Crumpet	Erica	36	2012-09-27 11:30:00

Antwort:

 select mems . surname , mems . firstname , mems . memid , min ( bks . starttime ) as starttime
	from cd . bookings bks
	inner join cd . members mems on
		mems . memid = bks . memid
	where starttime >= ' 2012-09-01 '
	group by mems . surname , mems . firstname , mems . memid
order by mems . memid ;          

This answer demonstrates the use of aggregate functions on dates. MIN works exactly as you'd expect, pulling out the lowest possible date in the result set. To make this work, we need to ensure that the result set only contains dates from September onwards. We do this using the WHERE clause.

You might typically use a query like this to find a customer's next booking. You can use this by replacing the date '2012-09-01' with the function now()

Produce a list of member names, with each row containing the total member count

Produce a list of member names, with each row containing the total member count. Order by join date.

Expected results:

zählen	firstname	Nachname
31	GAST	GAST
31	Darren	Schmied
31	Tracy	Schmied
31	Tim	Rownam
31	Janice	Joplette
31	Gerald	Butters
31	Burton	Tracy
31	Nancy	Wagen
31	Tim	Boothe
31	Ponder	Stibbons
31	Charles	Owen
31	David	Jones
31	Anne	Bäcker
31	Jemima	Farrell
31	Jack	Schmied
31	Florenz	Bader
31	Timothy	Bäcker
31	David	Pinker
31	Matthew	Genting
31	Anna	Mackenzie
31	Joan	Coplin
31	Ramnaresh	Sarwin
31	Douglas	Jones
31	Henrietta	Rumney
31	David	Farrell
31	Henry	Worthington-Smyth
31	Millicent	Aufgabe
31	Hyazinthe	Tupperware
31	John	Jagd
31	Erica	Crumpet
31	Darren	Schmied

Antwort:

 select count ( * ) over(), firstname, surname
	from cd . members
order by joindate          

Using the knowledge we've built up so far, the most obvious answer to this is below. We use a subquery because otherwise SQL will require us to group by firstname and surname, producing a different result to what we're looking for.

 select ( select count ( * ) from cd . members ) as count, firstname, surname
	from cd . members
order by joindate

There's nothing at all wrong with this answer, but we've chosen a different approach to introduce a new concept called window functions. Window functions provide enormously powerful capabilities, in a form often more convenient than the standard aggregation functions. While this exercise is only a toy, we'll be working on more complicated examples in the near future.

Window functions operate on the result set of your (sub-)query, after the WHERE clause and all standard aggregation. They operate on a window of data. By default this is unrestricted: the entire result set, but it can be restricted to provide more useful results. For example, suppose instead of wanting the count of all members, we want the count of all members who joined in the same month as that member:

 select count ( * ) over(partition by date_trunc( ' month ' ,joindate)),
	firstname, surname
	from cd . members
order by joindate

In this example, we partition the data by month. For each row the window function operates over, the window is any rows that have a joindate in the same month. The window function thus produces a count of the number of members who joined in that month.

You can go further. Imagine if, instead of the total number of members who joined that month, you want to know what number joinee they were that month. You can do this by adding in an ORDER BY to the window function:

 select count ( * ) over(partition by date_trunc( ' month ' ,joindate) order by joindate),
	firstname, surname
	from cd . members
order by joindate

The ORDER BY changes the window again. Instead of the window for each row being the entire partition, the window goes from the start of the partition to the current row, and not beyond. Thus, for the first member who joins in a given month, the count is 1. For the second, the count is 2, and so on.

One final thing that's worth mentioning about window functions: you can have multiple unrelated ones in the same query. Try out the query below for an example - you'll see the numbers for the members going in opposite directions! This flexibility can lead to more concise, readable, and maintainable queries.

 select count ( * ) over(partition by date_trunc( ' month ' ,joindate) order by joindate asc ), 
	count ( * ) over(partition by date_trunc( ' month ' ,joindate) order by joindate desc ), 
	firstname, surname
	from cd . members
order by joindate

Window functions are extraordinarily powerful, and they will change the way you write and think about SQL. Make good use of them!

Produce a numbered list of members

Produce a monotonically increasing numbered list of members, ordered by their date of joining. Remember that member IDs are not guaranteed to be sequential.

Expected results:

row_number	firstname	Nachname
1	GAST	GAST
2	Darren	Schmied
3	Tracy	Schmied
4	Tim	Rownam
5	Janice	Joplette
6	Gerald	Butters
7	Burton	Tracy
8	Nancy	Wagen
9	Tim	Boothe
10	Ponder	Stibbons
11	Charles	Owen
12	David	Jones
13	Anne	Bäcker
14	Jemima	Farrell
15	Jack	Schmied
16	Florenz	Bader
17	Timothy	Bäcker
18	David	Pinker
19	Matthew	Genting
20	Anna	Mackenzie
21	Joan	Coplin
22	Ramnaresh	Sarwin
23	Douglas	Jones
24	Henrietta	Rumney
25	David	Farrell
26	Henry	Worthington-Smyth
27	Millicent	Aufgabe
28	Hyazinthe	Tupperware
29	John	Jagd
30	Erica	Crumpet
31	Darren	Schmied

Antwort:

 select row_number() over( order by joindate), firstname, surname
	from cd . members
order by joindate          

This exercise is a simple bit of window function practise! You could just as easily use count(*) over(order by joindate) here, so don't worry if you used that instead.

In this query, we don't define a partition, meaning that the partition is the entire dataset. Since we define an order for the window function, for any given row the window is: start of the dataset -> current row.

Output the facility id that has the highest number of slots booked, again

Output the facility id that has the highest number of slots booked. Ensure that in the event of a tie, all tieing results get output.

Expected results:

facid	gesamt
4	1404

Antwort:

 select facid, total from (
	select facid, sum (slots) total, rank() over ( order by sum (slots) desc ) rank
        	from cd . bookings
		group by facid
	) as ranked
	where rank = 1          

You may recall that this is a problem we've already solved in an earlier exercise. We came up with an answer something like below, which we then cut down using CTEs:

 select facid, sum (slots) as totalslots
	from cd . bookings
	group by facid
	having sum (slots) = ( select max ( sum2 . totalslots ) from
		( select sum (slots) as totalslots
		from cd . bookings
		group by facid
		) as sum2);

Once we've cleaned it up, this solution is perfectly adequate. Explaining how the query works makes it seem a little odd, though - 'find the number of slots booked by the best facility. Calculate the total slots booked for each facility, and return only the rows where the slots booked are the same as for the best'. Wouldn't it be nicer to be able to say 'calculate the number of slots booked for each facility, rank them, and pick out any at rank 1'?

Fortunately, window functions allow us to do this - although it's fair to say that doing so is not trivial to the untrained eye. The first key piece of information is the existence of the éfunction. This ranks values based on the ORDER BY that is passed to it. If there's a tie for (say) second place), the next gets ranked at position 4. So, what we need to do is get the number of slots for each facility, rank them, and pick off the ones at the top rank. A first pass at this might look something like the below:

 select facid, total from (
	select facid, total, rank() over ( order by total desc ) rank from (
		select facid, sum (slots) total
			from cd . bookings
			group by facid
		) as sumslots
	) as ranked
where rank = 1

The inner query calculates the total slots booked, the middle one ranks them, and the outer one creams off the top ranked. We can actually tidy this up a little: recall that window function get applied pretty late in the select function, after aggregation. That being the case, we can move the aggregation into the ORDER BY part of the function, as shown in the approved answer.

While the window function approach isn't massively simpler in terms of lines of code, it arguably makes more semantic sense.

Rank members by (rounded) hours used

Produce a list of members, along with the number of hours they've booked in facilities, rounded to the nearest ten hours. Rank them by this rounded figure, producing output of first name, surname, rounded hours, rank. Sort by rank, surname, and first name.

Expected results:

firstname	Nachname	Std.	Rang
GAST	GAST	1200	1
Darren	Schmied	340	2
Tim	Rownam	330	3
Tim	Boothe	220	4
Tracy	Schmied	220	4
Gerald	Butters	210	6
Burton	Tracy	180	7
Charles	Owen	170	8
Janice	Joplette	160	9
Anne	Bäcker	150	10
Timothy	Bäcker	150	10
David	Jones	150	10
Nancy	Wagen	130	13
Florenz	Bader	120	14
Anna	Mackenzie	120	14
Ponder	Stibbons	120	14
Jack	Schmied	110	17
Jemima	Farrell	90	18
David	Pinker	80	19
Ramnaresh	Sarwin	80	19
Matthew	Genting	70	21
Joan	Coplin	50	22
David	Farrell	30	23
Henry	Worthington-Smyth	30	23
John	Jagd	20	25
Douglas	Jones	20	25
Millicent	Aufgabe	20	25
Henrietta	Rumney	20	25
Erica	Crumpet	10	29
Hyazinthe	Tupperware	10	29

Antwort:

 select firstname, surname,
	(( sum ( bks . slots ) + 10 ) / 20 ) * 10 as hours,
	rank() over ( order by (( sum ( bks . slots ) + 10 ) / 20 ) * 10 desc ) as rank

	from cd . bookings bks
	inner join cd . members mems
		on bks . memid = mems . memid
	group by mems . memid
order by rank, surname, firstname;          

This answer isn't a great stretch over our previous exercise, although it does illustrate the function of RANK better. You can see that some of the clubgoers have an equal rounded number of hours booked in, and their rank is the same. If position 2 is shared between two members, the next one along gets position 4. There's a different function, DENSE_RANK , that would assign that member position 3 instead.

It's worth noting the technique we use to do rounding here. Adding 5, dividing by 10, and multiplying by 10 has the effect (thanks to integer arithmetic cutting off fractions) of rounding a number to the nearest 10. In our case, because slots are half an hour, we need to add 10, divide by 20, and multiply by 10. One could certainly make the argument that we should do the slots -> hours conversion independently of the rounding, which would increase clarity.

Talking of clarity, this rounding malarky is starting to introduce a noticeable amount of code repetition. At this point it's a judgement call, but you may wish to factor it out using a subquery as below:

 select firstname, surname, hours, rank() over ( order by hours desc ) from
	( select firstname, surname,
		(( sum ( bks . slots ) + 10 ) / 20 ) * 10 as hours

		from cd . bookings bks
		inner join cd . members mems
			on bks . memid = mems . memid
		group by mems . memid
	) as subq
order by rank, surname, firstname;

Find the top three revenue generating facilities

Produce a list of the top three revenue generating facilities (including ties). Output facility name and rank, sorted by rank and facility name.

Expected results:

Name	Rang
Massage Room 1	1
Massage Room 2	2
Tennis Court 2	3

Antwort:

 select name, rank from (
	select facs . name as name, rank() over ( order by sum (case
				when memid = 0 then slots * facs . guestcost
				else slots * membercost
			end) desc ) as rank
		from cd . bookings bks
		inner join cd . facilities facs
			on bks . facid = facs . facid
		group by facs . name
	) as subq
	where rank <= 3
order by rank;          

This question doesn't introduce any new concepts, and is just intended to give you the opportunity to practise what you already know. We use the CASE statement to calculate the revenue for each slot, and aggregate that on a per-facility basis using SUM . We then use the RANK window function to produce a ranking, wrap it all up in a subquery, and extract everything with a rank less than or equal to 3.

Classify facilities by value

Classify facilities into equally sized groups of high, average, and low based on their revenue. Order by classification and facility name.

Expected results:

Name	Einnahmen
Massage Room 1	hoch
Massage Room 2	hoch
Tennis Court 2	hoch
Badminton Court	Durchschnitt
Squash Court	Durchschnitt
Tennis Court 1	Durchschnitt
Pool Table	niedrig
Snooker Table	niedrig
Tischtennis	niedrig

Antwort:

 select name, case when class = 1 then ' high '
		when class = 2 then ' average '
		else ' low '
		end revenue
	from (
		select facs . name as name, ntile( 3 ) over ( order by sum (case
				when memid = 0 then slots * facs . guestcost
				else slots * membercost
			end) desc ) as class
		from cd . bookings bks
		inner join cd . facilities facs
			on bks . facid = facs . facid
		group by facs . name
	) as subq
order by class, name;          

This exercise should mostly use familiar concepts, although we do introduce the NTILE window function. NTILE groups values into a passed-in number of groups, as evenly as possible. It outputs a number from 1->number of groups. We then use a CASE statement to turn that number into a label!

Calculate the payback time for each facility

Based on the 3 complete months of data so far, calculate the amount of time each facility will take to repay its cost of ownership. Remember to take into account ongoing monthly maintenance. Output facility name and payback time in months, order by facility name. Don't worry about differences in month lengths, we're only looking for a rough value here!

Expected results:

Name	Monate
Badminton Court	6.8317677198975235
Massage Room 1	0.18885741265344664778
Massage Room 2	1.7621145374449339
Pool Table	5.3333333333333333
Snooker Table	6.9230769230769231
Squash Court	1.1339582703356516
Tischtennis	6.4000000000000000
Tennis Court 1	2.2624434389140271
Tennis Court 2	1.7505470459518600

Antwort:

 select 	facs . name as name,
	facs . initialoutlay / (( sum (case
			when memid = 0 then slots * facs . guestcost
			else slots * membercost
		end) / 3 ) - facs . monthlymaintenance ) as months
	from cd . bookings bks
	inner join cd . facilities facs
		on bks . facid = facs . facid
	group by facs . facid
order by name;          

In contrast to all our recent exercises, there's no need to use window functions to solve this problem: it's just a bit of maths involving monthly revenue, initial outlay, and monthly maintenance. Again, for production code you might want to clarify what's going on a little here using a subquery (although since we've hard-coded the number of months, putting this into production is unlikely!). A tidied-up version might look like:

 select 	name, 
	initialoutlay / (monthlyrevenue - monthlymaintenance) as repaytime 
	from 
		( select facs . name as name, 
			facs . initialoutlay as initialoutlay,
			facs . monthlymaintenance as monthlymaintenance,
			sum (case
				when memid = 0 then slots * facs . guestcost
				else slots * membercost
			end) / 3 as monthlyrevenue
		from cd . bookings bks
		inner join cd . facilities facs
			on bks . facid = facs . facid
		group by facs . facid
	) as subq
order by name;

But, I hear you ask, what would an automatic version of this look like? One that didn't need to have a hard-coded number of months in it? That's a little more complicated, and involves some date arithmetic. I've factored that out into a CTE to make it a little more clear.

with monthdata as (
	select 	mincompletemonth,
		maxcompletemonth,
		(extract(year from maxcompletemonth) * 12 ) +
			extract(month from maxcompletemonth) -
			(extract(year from mincompletemonth) * 12 ) -
			extract(month from mincompletemonth) as nummonths 
	from (
		select 	date_trunc( ' month ' , 
				( select max (starttime) from cd . bookings )) as maxcompletemonth,
			date_trunc( ' month ' , 
				( select min (starttime) from cd . bookings )) as mincompletemonth
	) as subq
)
select 	name, 
	initialoutlay / (monthlyrevenue - monthlymaintenance) as repaytime 
	
	from
		( select facs . name as name,
			facs . initialoutlay as initialoutlay,
			facs . monthlymaintenance as monthlymaintenance,
			sum (case
				when memid = 0 then slots * facs . guestcost
				else slots * membercost
			end) / ( select nummonths from monthdata) as monthlyrevenue
			
			from cd . bookings bks
			inner join cd . facilities facs
				on bks . facid = facs . facid
			where bks . starttime < ( select maxcompletemonth from monthdata)
			group by facs . facid
		) as subq
order by name;

This code restricts the data that goes in to complete months. It does this by selecting the maximum date, rounding down to the month, and stripping out all dates larger than that. Even this code is not completely-complete. It doesn't handle the case of a facility making a loss. Fixing that is not too hard, and is left as (another) exercise for the reader!

Calculate a rolling average of total revenue

For each day in August 2012, calculate a rolling average of total revenue over the previous 15 days. Output should contain date and revenue columns, sorted by the date. Remember to account for the possibility of a day having zero revenue. This one's a bit tough, so don't be afraid to check out the hint!

Expected results:

Datum	Einnahmen
2012-08-01	1126.8333333333333333
2012-08-02	1153.0000000000000000
2012-08-03	1162.9000000000000000
2012-08-04	1177.3666666666666667
2012-08-05	1160.9333333333333333
2012-08-06	1185.4000000000000000
2012-08-07	1182.8666666666666667
2012-08-08	1172.6000000000000000
2012-08-09	1152.4666666666666667
2012-08-10	1175.0333333333333333
2012-08-11	1176.6333333333333333
2012-08-12	1195.6666666666666667
2012-08-13	1218.0000000000000000
2012-08-14	1247.4666666666666667
2012-08-15	1274.1000000000000000
2012-08-16	1281.2333333333333333
2012-08-17	1324.4666666666666667
2012-08-18	1373.7333333333333333
2012-08-19	1406.0666666666666667
2012-08-20	1427.0666666666666667
2012-08-21	1450.3333333333333333
2012-08-22	1539.7000000000000000
2012-08-23	1567.3000000000000000
2012-08-24	1592.3333333333333333
2012-08-25	1615.0333333333333333
2012-08-26	1631.2000000000000000
2012-08-27	1659.4333333333333333
2012-08-28	1687.0000000000000000
2012-08-29	1684.6333333333333333
2012-08-30	1657.9333333333333333
2012-08-31	1703.4000000000000000

Antwort:

 select 	dategen . date ,
	(
		-- correlated subquery that, for each day fed into it,
		-- finds the average revenue for the last 15 days
		select sum (case
			when memid = 0 then slots * facs . guestcost
			else slots * membercost
		end) as rev

		from cd . bookings bks
		inner join cd . facilities facs
			on bks . facid = facs . facid
		where bks . starttime > dategen . date - interval ' 14 days '
			and bks . starttime < dategen . date + interval ' 1 day '
	) / 15 as revenue
	from
	(
		-- generates a list of days in august
		select 	cast(generate_series( timestamp ' 2012-08-01 ' ,
			' 2012-08-31 ' , ' 1 day ' ) as date ) as date
	)  as dategen
order by dategen . date ;          

There's at least two equally good solutions to this question. I've put the simplest to write as the answer, but there's also a more flexible solution that uses window functions.

Let's look at the selected answer first. When I read SQL queries, I tend to read the SELECT part of the statement last - the FROM and WHERE parts tend to be more interesting. So, what do we have in our FROM ? A call to the GENERATE_SERIES function. This does pretty much what it says on the tin - generates a series of values. You can specify a start value, a stop value, and an increment. It works for integer types and dates - although, as you can see, we need to be explicit about what types are going into and out of the function. Try removing the casts, and seeing the result!

So, we've generated a timestamp for each day in August. Now, for each day, we need to generate our average. We can do this using a correlated subquery . If you remember, a correlated subquery is a subquery that uses values from the outer query. This means that it gets executed once for each result row in the outer query. This is in contrast to an uncorrelated subquery, which only has to be executed once.

If we look at our correlated subquery, we can see that it's correlated on the dategen.date field. It produces a sum of revenue for this day and the 14 days prior to it, and then divides that sum by 15. This produces the output we're looking for!

I mentioned that there's a window function-based solution for this problem as well - you can see it below. The approach we use for this is generating a list of revenue for each day, and then using window function aggregation over that list. The nice thing about this method is that once you have the per-day revenue, you can produce a wide range of results quite easily - you might, for example, want rolling averages for the previous month, 15 days, and 5 days. This is easy to do using this method, and rather harder using conventional aggregation.

 select date , avgrev from (
	-- AVG over this row and the 14 rows before it.
	select 	dategen . date as date ,
		avg ( revdata . rev ) over( order by dategen . date rows 14 preceding) as avgrev
	from
		-- generate a list of days.  This ensures that a row gets generated
		-- even if the day has 0 revenue.  Note that we generate days before
		-- the start of october - this is because our window function needs
		-- to know the revenue for those days for its calculations.
		( select
			cast(generate_series( timestamp ' 2012-07-10 ' , ' 2012-08-31 ' , ' 1 day ' ) as date ) as date
		)  as dategen
		left outer join
			-- left join to a table of per-day revenue
			( select cast( bks . starttime as date ) as date ,
				sum (case
					when memid = 0 then slots * facs . guestcost
					else slots * membercost
				end) as rev

				from cd . bookings bks
				inner join cd . facilities facs
					on bks . facid = facs . facid
				group by cast( bks . starttime as date )
			) as revdata
			on dategen . date = revdata . date
	) as subq
	where date >= ' 2012-08-01 '
order by date ;

You'll note that we've been wanting to work out daily revenue quite frequently. Rather than inserting that calculation into all our queries, which is rather messy (and will cause us a big headache if we ever change our schema), we probably want to store that information somewhere. Your first thought might be to calculate information and store it somewhere for later use. This is a common tactic for large data warehouses, but it can cause us some problems - if we ever go back and edit our data, we need to remember to recalculate. For non-enormous-scale data like we're looking at here, we can just create a view instead. A view is essentially a stored query that looks exactly like a table. Under the covers, the DBMS just subsititutes in the relevant portion of the view definition when you select data from it. They're very easy to create, as you can see below:

 create or replace view cd .dailyrevenue as
	select 	cast( bks . starttime as date ) as date ,
		sum (case
			when memid = 0 then slots * facs . guestcost
			else slots * membercost
		end) as rev

		from cd . bookings bks
		inner join cd . facilities facs
			on bks . facid = facs . facid
		group by cast( bks . starttime as date );

You can see that this makes our query an awful lot simpler!

 select date , avgrev from (
	select  dategen . date as date ,
		avg ( revdata . rev ) over( order by dategen . date rows 14 preceding) as avgrev
	from		
		( select
			cast(generate_series( timestamp ' 2012-07-10 ' , ' 2012-08-31 ' , ' 1 day ' ) as date ) as date
		)  as dategen
		left outer join
			cd . dailyrevenue as revdata on dategen . date = revdata . date
		) as subq
	where date >= ' 2012-08-01 '
order by date ;

As well as storing frequently-used query fragments, views can be used for a variety of purposes, including restricting access to certain columns of a table.

---

Working with Timestamps

Dates/Times in SQL are a complex topic, deserving of a category of their own. They're also fantastically powerful, making it easier to work with variable-length concepts like 'months' than many programming languages.

Before getting started on this category, it's probably worth taking a look over the PostgreSQL docs page on date/time functions. You might also want to complete the aggregate functions category, since we'll use some of those capabilities in this section.

Produce a timestamp for 1 am on the 31st of August 2012

Produce a timestamp for 1 am on the 31st of August 2012.

Expected results:

timestamp
2012-08-31 01:00:00

Antwort:

 select timestamp ' 2012-08-31 01:00:00 ' ;          

Here's a pretty easy question to start off with! SQL has a bunch of different date and time types, which you can peruse at your leisure over at the excellent Postgres documentation. These basically allow you to store dates, times, or timestamps (date+time).

The approved answer is the best way to create a timestamp under normal circumstances. You can also use casts to change a correctly formatted string into a timestamp, for example:

 select ' 2012-08-31 01:00:00 ' :: timestamp ;
select cast( ' 2012-08-31 01:00:00 ' as timestamp );

The former approach is a Postgres extension, while the latter is SQL-standard. You'll note that in many of our earlier questions, we've used bare strings without specifying a data type. This works because when Postgres is working with a value coming out of a timestamp column of a table (say), it knows to cast our strings to timestamps.

Timestamps can be stored with or without time zone information. We've chosen not to here, but if you like you could format the timestamp like "2012-08-31 01:00:00 +00:00", assuming UTC. Note that timestamp with time zone is a different type to timestamp - when you're declaring it, you should use TIMESTAMP WITH TIME ZONE 2012-08-31 01:00:00 +00:00.

Finally, have a bit of a play around with some of the different date/time serialisations described in the Postgres docs. You'll find that Postgres is extremely flexible with the formats it accepts, although my recommendation to you would be to use the standard serialisation we've used here - you'll find it unambiguous and easy to port to other DBs.

Subtract timestamps from each other

Find the result of subtracting the timestamp '2012-07-30 01:00:00' from the timestamp '2012-08-31 01:00:00'

Expected results:

Intervall
32 days

Antwort:

 select timestamp ' 2012-08-31 01:00:00 ' - timestamp ' 2012-07-30 01:00:00 ' as interval;

Subtracting timestamps produces an INTERVAL data type. INTERVAL s are a special data type for representing the difference between two TIMESTAMP types. When subtracting timestamps, Postgres will typically give an interval in terms of days, hours, minutes, seconds, without venturing into months. This generally makes life easier, since months are of variable lengths.

One of the useful things about intervals, though, is the fact that they can encode months. Let's imagine that I want to schedule something to occur in exactly one month's time, regardless of the length of my month. To do this, I could use [timestamp] + interval '1 month' .

Intervals stand in contrast to SQL's treatment of DATE types. Dates don't use intervals - instead, subtracting two dates will return an integer representing the number of days between the two dates. You can also add integer values to dates. This is sometimes more convenient, depending on how much intelligence you require in the handling of your dates!

Generate a list of all the dates in October 2012

Produce a list of all the dates in October 2012. They can be output as a timestamp (with time set to midnight) or a date.

Expected results:

ts
2012-10-01 00:00:00
2012-10-02 00:00:00
2012-10-03 00:00:00
2012-10-04 00:00:00
2012-10-05 00:00:00
2012-10-06 00:00:00
2012-10-07 00:00:00
2012-10-08 00:00:00
2012-10-09 00:00:00
2012-10-10 00:00:00
2012-10-11 00:00:00
2012-10-12 00:00:00
2012-10-13 00:00:00
2012-10-14 00:00:00
2012-10-15 00:00:00
2012-10-16 00:00:00
2012-10-17 00:00:00
2012-10-18 00:00:00
2012-10-19 00:00:00
2012-10-20 00:00:00
2012-10-21 00:00:00
2012-10-22 00:00:00
2012-10-23 00:00:00
2012-10-24 00:00:00
2012-10-25 00:00:00
2012-10-26 00:00:00
2012-10-27 00:00:00
2012-10-28 00:00:00
2012-10-29 00:00:00
2012-10-30 00:00:00
2012-10-31 00:00:00

Antwort:

 select generate_series( timestamp ' 2012-10-01 ' , timestamp ' 2012-10-31 ' , interval ' 1 day ' ) as ts;          

One of the best features of Postgres over other DBs is a simple function called GENERATE_SERIES . This function allows you to generate a list of dates or numbers, specifying a start, an end, and an increment value. It's extremely useful for situations where you want to output, say, sales per day over the course of a month. A typical way to do that on a table containing a list of sales might be to use a SUM aggregation, grouping by the date and product type. Unfortunately, this approach has a flaw: if there are no sales for a given day, it won't show up! To make it work properly, you need to left join from a sequential list of timestamps to the aggregated data to fill in the blank spaces.

On other database systems, it's not uncommon to keep a 'calendar table' full of dates, with which you can perform these joins. Alternatively, on some systems you can write an analogue to generate_series using recursive CTEs. Fortunately for us, Postgres makes our lives a lot easier!

Get the day of the month from a timestamp

Get the day of the month from the timestamp '2012-08-31' as an integer.

Expected results:

date_part
31

Antwort:

 select extract(day from timestamp ' 2012-08-31 ' );          

The EXTRACT function is used for getting sections of a timestamp or interval. You can get the value of any field in the timestamp as an integer.

Work out the number of seconds between timestamps

Work out the number of seconds between the timestamps '2012-08-31 01:00:00' and '2012-09-02 00:00:00'

Expected results:

date_part
169200

Antwort:

 select extract(epoch from ( timestamp ' 2012-09-02 00:00:00 ' - ' 2012-08-31 01:00:00 ' ));          

The above answer is a Postgres-specific trick. Extracting the epoch converts an interval or timestamp into a number of seconds, or the number of seconds since epoch (January 1st, 1970) respectively. If you want the number of minutes, hours, etc you can just divide the number of seconds appropriately.

If you want to write more portable code, you will unfortunately find that you cannot use extract epoch . Instead you will need to use something like:

 select 	extract(day from ts . int ) * 60 * 60 * 24 +
	extract(hour from ts . int ) * 60 * 60 + 
	extract(minute from ts . int ) * 60 +
	extract(second from ts . int )
	from
		( select timestamp ' 2012-09-02 00:00:00 ' - ' 2012-08-31 01:00:00 ' as int ) ts

Antwort:

This is, as you can observe, rather awful. If you're planning to write cross platform SQL, I would consider having a library of common user defined functions for each DBMS, allowing you to normalise any common requirements like this. This keeps your main codebase a lot cleaner.

Work out the number of days in each month of 2012

For each month of the year in 2012, output the number of days in that month. Format the output as an integer column containing the month of the year, and a second column containing an interval data type.

Expected results:

Monat	Länge
1	31 days
2	29 days
3	31 days
4	30 Tage
5	31 days
6	30 Tage
7	31 days
8	31 days
9	30 Tage
10	31 days
11	30 Tage
12	31 days

Antwort:

 select 	extract(month from cal . month ) as month,
	( cal . month + interval ' 1 month ' ) - cal . month as length
	from
	(
		select generate_series( timestamp ' 2012-01-01 ' , timestamp ' 2012-12-01 ' , interval ' 1 month ' ) as month
	) cal
order by month;          

This answer shows several of the concepts we've learned. We use the GENERATE_SERIES function to produce a year's worth of timestamps, incrementing a month at a time. We then use the EXTRACT function to get the month number. Finally, we subtract each timestamp + 1 month from itself.

It's worth noting that subtracting two timestamps will always produce an interval in terms of days (or portions of a day). You won't just get an answer in terms of months or years, because the length of those time periods is variable.

Work out the number of days remaining in the month

For any given timestamp, work out the number of days remaining in the month. The current day should count as a whole day, regardless of the time. Use '2012-02-11 01:00:00' as an example timestamp for the purposes of making the answer. Format the output as a single interval value.

Expected results:

übrig
19 days

Antwort:

 select (date_trunc( ' month ' , ts . testts ) + interval ' 1 month ' ) 
		- date_trunc( ' day ' , ts . testts ) as remaining
	from ( select timestamp ' 2012-02-11 01:00:00 ' as testts) ts          

The star of this particular show is the DATE_TRUNC function. It does pretty much what you'd expect - truncates a date to a given minute, hour, day, month, and so on. The way we've solved this problem is to truncate our timestamp to find the month we're in, add a month to that, and subtract our timestamp. To ensure partial days get treated as whole days, the timestamp we subtract is truncated to the nearest day.

Note the way we've put the timestamp into a subquery. This isn't required, but it does mean you can give the timestamp a name, rather than having to list the literal repeatedly.

Work out the end time of bookings

Return a list of the start and end time of the last 10 bookings (ordered by the time at which they end, followed by the time at which they start) in the system.

Expected results:

starttime	endtime
2013-01-01 15:30:00	2013-01-01 16:00:00
2012-09-30 19:30:00	2012-09-30 20:30:00
2012-09-30 19:00:00	2012-09-30 20:30:00
2012-09-30 19:30:00	2012-09-30 20:00:00
2012-09-30 19:00:00	2012-09-30 20:00:00
2012-09-30 19:00:00	2012-09-30 20:00:00
2012-09-30 18:30:00	2012-09-30 20:00:00
2012-09-30 18:30:00	2012-09-30 20:00:00
2012-09-30 19:00:00	2012-09-30 19:30:00
2012-09-30 18:30:00	2012-09-30 19:30:00

Antwort:

 select starttime, starttime + slots * (interval ' 30 minutes ' ) endtime
	from cd . bookings
	order by endtime desc , starttime desc
	limit 10          

This question simply returns the start time for a booking, and a calculated end time which is equal to start time + (30 minutes * slots) . Note that it's perfectly okay to multiply intervals.

The other thing you'll notice is the use of order by and limit to get the last ten bookings. All this does is order the bookings by the (descending) time at which they end, and pick off the top ten.

Return a count of bookings for each month

Return a count of bookings for each month, sorted by month

Expected results:

Monat	zählen
2012-07-01 00:00:00	658
2012-08-01 00:00:00	1472
2012-09-01 00:00:00	1913
2013-01-01 00:00:00	1

Antwort:

 select date_trunc( ' month ' , starttime) as month, count ( * )
	from cd . bookings
	group by month
	order by month          

This one is a fairly simple reuse of concepts we've seen before. We simply count the number of bookings, and aggregate by the booking's start time, truncated to the month.

Work out the utilisation percentage for each facility by month

Work out the utilisation percentage for each facility by month, sorted by name and month, rounded to 1 decimal place. Opening time is 8am, closing time is 8.30pm. You can treat every month as a full month, regardless of if there were some dates the club was not open.

Expected results:

Name	Monat	Verwendung
Badminton Court	2012-07-01 00:00:00	23.2
Badminton Court	2012-08-01 00:00:00	59,2
Badminton Court	2012-09-01 00:00:00	76.0
Massage Room 1	2012-07-01 00:00:00	34.1
Massage Room 1	2012-08-01 00:00:00	63,5
Massage Room 1	2012-09-01 00:00:00	86.4
Massage Room 2	2012-07-01 00:00:00	3.1
Massage Room 2	2012-08-01 00:00:00	10.6
Massage Room 2	2012-09-01 00:00:00	16.3
Pool Table	2012-07-01 00:00:00	15.1
Pool Table	2012-08-01 00:00:00	41,5
Pool Table	2012-09-01 00:00:00	62,8
Pool Table	2013-01-01 00:00:00	0.1
Snooker Table	2012-07-01 00:00:00	20.1
Snooker Table	2012-08-01 00:00:00	42.1
Snooker Table	2012-09-01 00:00:00	56,8
Squash Court	2012-07-01 00:00:00	21.2
Squash Court	2012-08-01 00:00:00	51.6
Squash Court	2012-09-01 00:00:00	72.0
Tischtennis	2012-07-01 00:00:00	13.4
Tischtennis	2012-08-01 00:00:00	39.2
Tischtennis	2012-09-01 00:00:00	56.3
Tennis Court 1	2012-07-01 00:00:00	34.8
Tennis Court 1	2012-08-01 00:00:00	59,2
Tennis Court 1	2012-09-01 00:00:00	78.8
Tennis Court 2	2012-07-01 00:00:00	26.7
Tennis Court 2	2012-08-01 00:00:00	62.3
Tennis Court 2	2012-09-01 00:00:00	78.4

Antwort:

 select name, month, 
	round(( 100 * slots) /
		cast(
			25 * (cast((month + interval ' 1 month ' ) as date )
			- cast (month as date )) as numeric ), 1 ) as utilisation
	from  (
		select facs . name as name, date_trunc( ' month ' , starttime) as month, sum (slots) as slots
			from cd . bookings bks
			inner join cd . facilities facs
				on bks . facid = facs . facid
			group by facs . facid , month
	) as inn
order by name, month          

The meat of this query (the inner subquery) is really quite simple: an aggregation to work out the total number of slots used per facility per month. If you've covered the rest of this section and the category on aggregates, you likely didn't find this bit too challenging.

This query does, unfortunately, have some other complexity in it: working out the number of days in each month. We can calculate the number of days between two months by subtracting two timestamps with a month between them. This, unfortunately, gives us back on interval datatype, which we can't use to do mathematics. In this case we've worked around that limitation by converting our timestamps into dates before subtracting. Subtracting date types gives us an integer number of days.

A alternative to this workaround is to convert the interval into an epoch value: that is, a number of seconds. To do this use EXTRACT(EPOCH FROM month)/(24*60*60) . This is arguably a much nicer way to do things, but is much less portable to other database systems.

---

String Operations

String operations in most RDBMSs are, arguably, needlessly painful. Fortunately, Postgres is better than most in this regard, providing strong regular expression support. This section covers basic string manipulation, use of the LIKE operator, and use of regular expressions. I also make an effort to show you some alternative approaches that work reliably in most RDBMSs. Be sure to check out Postgres' string function docs page if you're not confident about these exercises.

Anthony Molinaro's SQL Cookbook provides some excellent documentation of (difficult) cross-DBMS compliant SQL string manipulation. I'd strongly recommend his book, particularly as it's published by O'Reilly, whose ethical policy of DRM-free ebook distribution deserves rich rewards.

Format the names of members

Output the names of all members, formatted as 'Surname, Firstname'

Expected results:

Name
GUEST, GUEST
Smith, Darren
Smith, Tracy
Rownam, Tim
Joplette, Janice
Butters, Gerald
Tracy, Burton
Dare, Nancy
Boothe, Tim
Stibbons, Ponder
Owen, Charles
Jones, David
Baker, Anne
Farrell, Jemima
Smith, Jack
Bader, Florence
Baker, Timothy
Pinker, David
Genting, Matthew
Mackenzie, Anna
Coplin, Joan
Sarwin, Ramnaresh
Jones, Douglas
Rumney, Henrietta
Farrell, David
Worthington-Smyth, Henry
Purview, Millicent
Tupperware, Hyacinth
Hunt, John
Crumpet, Erica
Smith, Darren

Antwort:

 select surname || ' , ' || firstname as name from cd . members          

Building strings in sql is similar to other languages, with the exception of the concatenation operator: ||. Some systems (like SQL Server) use +, but || is the SQL standard.

Find facilities by a name prefix

Find all facilities whose name begins with 'Tennis'. Retrieve all columns.

Expected results:

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	5	25	10000	200
1	Tennis Court 2	5	25	8000	200

Antwort:

 select * from cd . facilities where name like ' Tennis% ' ;          

The SQL LIKE operator is a highly standard way of searching for a string using basic matching. The % character matches any string, while _ matches any single character.

One point that's worth considering when you use LIKE is how it uses indexes. If you're using the 'C' locale, any LIKE string with a fixed beginning (as in our example here) can use an index. If you're using any other locale, LIKE will not use any index by default. See here for details on how to change that.

Perform a case-insensitive search

Perform a case-insensitive search to find all facilities whose name begins with 'tennis'. Retrieve all columns.

Expected results:

facid	Name	membercost	guestcost	initialoutlay	monthlymaintenance
0	Tennis Court 1	5	25	10000	200
1	Tennis Court 2	5	25	8000	200

Antwort:

 select * from cd . facilities where upper (name) like ' TENNIS% ' ;          

There's no direct operator for case-insensitive comparison in standard SQL. Fortunately, we can take a page from many other language's books, and simply force all values into upper case when we do our comparison. This renders case irrelevant, and gives us our result.

Alternatively, Postgres does provide the ILIKE operator, which performs case insensitive searches. This isn't standard SQL, but it's arguably more clear.

You should realise that running a function like UPPER over a column value prevents Postgres from making use of any indexes on the column (the same is true for ILIKE ). Fortunately, Postgres has got your back: rather than simply creating indexes over columns, you can also create indexes over expressions. If you created an index over UPPER(name) , this query could use it quite happily.

Find telephone numbers with parentheses

You've noticed that the club's member table has telephone numbers with very inconsistent formatting. You'd like to find all the telephone numbers that contain parentheses, returning the member ID and telephone number sorted by member ID.

Expected results:

memid	Telefon
0	[phone]
3	[phone]
4	[phone]
5	[phone]
6	[phone]
7	[phone]
8	[phone]
9	[phone]
10	[phone]
11	[phone]
13	[phone]
14	[phone]
15	[phone]
20	[phone]
21	[phone]
22	[phone]
24	[phone]
27	[phone]
28	[phone]
29	[phone]
30	[phone]
33	[phone]
35	[phone]
36	[phone]
37	[phone]

Antwort:

 select memid, telephone from cd . members where telephone ~ ' [()] ' ;          

We've chosen to answer this using regular expressions, although Postgres does provide other string functions like POSITION that would do the job at least as well. Postgres implements POSIX regular expression matching via the ~ operator. If you've used regular expressions before, the functionality of the operator will be very familiar to you.

As an alternative, you can use the SQL standard SIMILAR TO operator. The regular expressions for this have similarities to the POSIX standard, but a lot of differences as well. Some of the most notable differences are:

• As in the LIKE operator, SIMILAR TO uses the '_' character to mean 'any character', and the '%' character to mean 'any string'.
• A SIMILAR TO expression must match the whole string, not just a substring as in posix regular expressions. This means that you'll typically end up bracketing an expression in '%' characters.
• Der '.' character does not mean 'any character' in SIMILAR TO regexes: it's just a plain character.

The SIMILAR TO equivalent of the given answer is shown below:

 select memid, telephone from cd . members where telephone similar to ' %[()]% ' ;

Finally, it's worth noting that regular expressions usually don't use indexes. Generally you don't want your regex to be responsible for doing heavy lifting in your query, because it will be slow. If you need fuzzy matching that works fast, consider working out if your needs can be met by full text search.

Pad zip codes with leading zeroes

The zip codes in our example dataset have had leading zeroes removed from them by virtue of being stored as a numeric type. Retrieve all zip codes from the members table, padding any zip codes less than 5 characters long with leading zeroes. Order by the new zip code.

Expected results:

Reißverschluss
00000
00234
00234
04321
04321
10383
11986
23423
28563
33862
34232
43532
43533
45678
52365
54333
56754
57392
58393
64577
65332
65464
66796
68666
69302
75655
78533
80743
84923
87630
97676

Antwort:

 select lpad(cast(zipcode as char ( 5 )), 5 , ' 0 ' ) zip from cd . members order by zip          

Postgres' LPAD function is the star of this particular show. It does basically what you'd expect: allow us to produce a padded string. We need to remember to cast the zipcode to a string for it to be accepted by the LPAD function.

When inheriting an old database, It's not that unusual to find wonky decisions having been made over data types. You may wish to fix mistakes like these, but have a lot of code that would break if you changed datatypes. In that case, one option (depending on performance requirements) is to create a view over your table which presents the data in a fixed-up manner, and gradually migrate.

Count the number of members whose surname starts with each letter of the alphabet

You'd like to produce a count of how many members you have whose surname starts with each letter of the alphabet. Sort by the letter, and don't worry about printing out a letter if the count is 0.

Expected results:

Brief	zählen
B	5
C	2
D	1
F	2
G	2
H	1
J	3
M	1
O	1
P	2
R	2
S	6
T	2
W	1

Antwort:

 select substr ( mems . surname , 1 , 1 ) as letter, count ( * ) as count 
    from cd . members mems
    group by letter
    order by letter          

This exercise is fairly straightforward. You simply need to retrieve the first letter of the member's surname, and do some basic aggregation to achieve a count. We use the SUBSTR function here, but there's a variety of other ways you can achieve the same thing. The LEFT function, for example, returns you the first n characters from the left of the string. Alternatively, you could use the SUBSTRING function, which allows you to use regular expressions to extract a portion of the string.

One point worth noting: as you can see, string functions in SQL are based on 1-indexing, not the 0-indexing that you're probably used to. This will likely trip you up once or twice before you get used to it :-)

Clean up telephone numbers

The telephone numbers in the database are very inconsistently formatted. You'd like to print a list of member ids and numbers that have had '-','(',')', and ' ' characters removed. Order by member id.

Expected results:

memid	Telefon
0	0000000000
1	5555555555
2	5555555555
3	8446930723
4	8339424710
5	8440784130
6	8223549973
7	8337764001
8	8114332547
9	8331603900
10	8555425251
11	8445368036
12	8440765141
13	8550160163
14	8221633254
15	8334993527
16	8339410824
17	8114096734
20	8119721377
21	8226612898
22	8224992232
24	8224131470
26	8445368036
27	8229898876
28	8557559876
29	8558943758
30	8559419786
33	8226655327
35	8997206978
36	8117324816
37	8225773541

Antwort:

 select memid, translate (telephone, ' -() ' , ' ' ) as telephone
    from cd . members
    order by memid;          

The most direct solution is probably the TRANSLATE function, which can be used to replace characters in a string. You pass it three strings: the value you want altered, the characters to replace, and the characters you want them replaced with. In our case, we want all the characters deleted, so our third parameter is an empty string.

As is often the way with strings, we can also use regular expressions to solve our problem. The REGEXP_REPLACE function provides what we're looking for: we simply pass a regex that matches all non-digit characters, and replace them with nothing, as shown below. The 'g' flag tells the function to replace as many instances of the pattern as it can find. This solution is perhaps more robust, as it cleans out more bad formatting.

 select memid, regexp_replace(telephone, ' [^0-9] ' , ' ' , ' g ' ) as telephone
    from cd . members
    order by memid;

Making automated use of free-formatted text data can be a chore. Ideally you want to avoid having to constantly write code to clean up the data before using it, so you should consider having your database enforce correct formatting for you. You can do this using a CHECK constraint on your column, which allow you to reject any poorly-formatted entry. It's tempting to perform this kind of validation in the application layer, and this is certainly a valid approach. As a general rule, if your database is getting used by multiple applications, favour pushing more of your checks down into the database to ensure consistent behaviour between the apps.

Occasionally, adding a constraint isn't feasible. You may, for example, have two different legacy applications asserting differently formatted information. If you're unable to alter the applications, you have a couple of options to consider. Firstly, you can define a trigger on your table. This allows you to intercept data before (or after) it gets asserted to your table, and normalise it into a single format. Alternatively, you could build a view over your table that cleans up information on the fly, as it's read out. Newer applications can read from the view and benefit from more reliably formatted information.

---

Recursive Queries

Common Table Expressions allow us to, effectively, create our own temporary tables for the duration of a query - they're largely a convenience to help us make more readable SQL. Using the WITH RECURSIVE modifier, however, it's possible for us to create recursive queries. This is enormously advantageous for working with tree and graph-structured data - imagine retrieving all of the relations of a graph node to a given depth, for example.

This category shows you some basic recursive queries that are possible using our dataset.

Find the upward recommendation chain for member ID 27

Find the upward recommendation chain for member ID 27: that is, the member who recommended them, and the member who recommended that member, and so on. Return member ID, first name, and surname. Order by descending member id.

Expected results:

recommender	firstname	Nachname
20	Matthew	Genting
5	Gerald	Butters
1	Darren	Schmied

Antwort:

with recursive recommenders(recommender) as (
	select recommendedby from cd . members where memid = 27
	union all
	select mems . recommendedby
		from recommenders recs
		inner join cd . members mems
			on mems . memid = recs . recommender
)
select recs . recommender , mems . firstname , mems . surname
	from recommenders recs
	inner join cd . members mems
		on recs . recommender = mems . memid
order by memid desc          

WITH RECURSIVE is a fantastically useful piece of functionality that many developers are unaware of. It allows you to perform queries over hierarchies of data, which is very difficult by other means in SQL. Such scenarios often leave developers resorting to multiple round trips to the database system.

You've seen WITH before. The Common Table Expressions (CTEs) defined by WITH give you the ability to produce inline views over your data. This is normally just a syntactic convenience, but the RECURSIVE modifier adds the ability to join against results already produced to produce even more. A recursive WITH takes the basic form of:

WITH RECURSIVE NAME(columns) as (
	< initial statement >
	UNION ALL 
	< recursive statement >
)

The initial statement populates the initial data, and then the recursive statement runs repeatedly to produce more. Each step of the recursion can access the CTE, but it sees within it only the data produced by the previous iteration. It repeats until an iteration produces no additional data.

The most simple example of a recursive WITH might look something like this:

with recursive increment(num) as (
	select 1
	union all
	select increment . num + 1 from increment where increment . num < 5
)
select * from increment;

The initial statement produces '1'. The first iteration of the recursive statement sees this as the content of increment , and produces '2'. The next iteration sees the content of increment as '2', and so on. Execution terminates when the recursive statement produces no additional data.

With the basics out of the way, it's fairly easy to explain our answer here. The initial statement gets the ID of the person who recommended the member we're interested in. The recursive statement takes the results of the initial statement, and finds the ID of the person who recommended them. This value gets forwarded on to the next iteration, and so on.

Now that we've constructed the recommenders CTE, all our main SELECT statement has to do is get the member IDs from recommenders, and join to them members table to find out their names.

Find the downward recommendation chain for member ID 1

Find the downward recommendation chain for member ID 1: that is, the members they recommended, the members those members recommended, and so on. Return member ID and name, and order by ascending member id.

Expected results:

memid	firstname	Nachname
4	Janice	Joplette
5	Gerald	Butters
7	Nancy	Wagen
10	Charles	Owen
11	David	Jones
14	Jack	Schmied
20	Matthew	Genting
21	Anna	Mackenzie
26	Douglas	Jones
27	Henrietta	Rumney

Antwort:

with recursive recommendeds(memid) as (
	select memid from cd . members where recommendedby = 1
	union all
	select mems . memid
		from recommendeds recs
		inner join cd . members mems
			on mems . recommendedby = recs . memid
)
select recs . memid , mems . firstname , mems . surname
	from recommendeds recs
	inner join cd . members mems
		on recs . memid = mems . memid
order by memid          

This is a pretty minor variation on the previous question. The essential difference is that we're now heading in the opposite direction. One interesting point to note is that unlike the previous example, this CTE produces multiple rows per iteration, by virtue of the fact that we're heading down the recommendation tree (following all branches) rather than up it.

Produce a CTE that can return the upward recommendation chain for any member

Produce a CTE that can return the upward recommendation chain for any member. You should be able to select recommender from recommenders where member=x. Demonstrate it by getting the chains for members 12 and 22. Results table should have member and recommender, ordered by member ascending, recommender descending.

Expected results:

Mitglied	recommender	firstname	Nachname
12	9	Ponder	Stibbons
12	6	Burton	Tracy
22	16	Timothy	Bäcker
22	13	Jemima	Farrell

Antwort:

with recursive recommenders(recommender, member) as (
	select recommendedby, memid
		from cd . members
	union all
	select mems . recommendedby , recs . member
		from recommenders recs
		inner join cd . members mems
			on mems . memid = recs . recommender
)
select recs . member member, recs . recommender , mems . firstname , mems . surname
	from recommenders recs
	inner join cd . members mems		
		on recs . recommender = mems . memid
	where recs . member = 22 or recs . member = 12
order by recs . member asc , recs . recommender desc          

This question requires us to produce a CTE that can calculate the upward recommendation chain for any user. Most of the complexity of working out the answer is in realising that we now need our CTE to produce two columns: one to contain the member we're asking about, and another to contain the members in their recommendation tree. Essentially what we're doing is producing a table that flattens out the recommendation hierarchy.

Since we're looking to produce the chain for every user, our initial statement needs to select data for each user: their ID and who recommended them. Subsequently, we want to pass the member field through each iteration without changing it, while getting the next recommender. You can see that the recursive part of our statement hasn't really changed, except to pass through the 'member' field.