postgresql exercises

PostgreSQLエクササイズ

これは、Alisdair OwenのPostgreSQL演習に関するすべての質問と回答の編集です。これらの問題を実際に解決すると、このガイドをスキミングするよりもさらに進むことができるので、訪問を必ず支払うようにしてください。

目次

• はじめる
  • 自分のPostgresシステムを使用したいです
  • スキーマ
• 簡単なSQLクエリ
  • テーブルからすべてを取得します
  • テーブルから特定の列を取得します
  • どの行が取得されるかを制御します
  • どの行を取得するかを制御、パート2
  • 基本的な文字列検索
  • 複数の可能な値と一致します
  • 結果をバケットに分類します
  • 日付を扱う
  • 重複を削除し、結果を注文します
  • 複数のクエリからの結果を組み合わせます
  • 単純な集約
  • より多くの集約
• 参加してサブクリーリー
  • メンバーの予約の開始時間を取得します
  • テニスコートの予約の開始時間を実現する
  • 別のメンバーを推薦したすべてのメンバーのリストを作成する
  • すべてのメンバーのリストを作成し、推奨
  • テニスコートを使用したすべてのメンバーのリストを作成する
  • 高価な予約のリストを作成します
  • すべてのメンバーのリストを作成し、参加者を使用しないでください
  • サブクエリを使用して、高価な予約のリストを作成します
• データの変更
  • いくつかのデータをテーブルに挿入します
  • データの複数の行をテーブルに挿入します
  • 計算されたデータをテーブルに挿入します
  • 既存のデータを更新します
  • 複数の行と列を同時に更新します
  • 別の行の内容に基づいて行を更新する
  • すべての予約を削除します
  • Cメンバーテーブルからメンバーを削除します
  • サブクエリに基づいて削除します
• 集約
  • 施設の数を数えます
  • 高価な施設の数を数えます
  • 各メンバーが作成する推奨事項の数を数えます
  • 施設ごとに予約された合計スロットをリストします
  • 特定の月に施設ごとに予約された合計スロットをリストします
  • 1か月あたりの施設ごとに予約された合計スロットをリストします
  • 少なくとも1つの予約をしたメンバーの数を見つける
  • 1000を超えるスロットが予約された設備をリストします
  • 各施設の総収益を見つけます
  • 総収益が1000未満の施設を見つけます
  • 予約されたスロットの数が最も多い施設IDを出力します
  • 1か月あたりの施設ごとに予約された合計スロット、パート2をリストします
  • 名前付き施設ごとに予約された合計時間をリストします
  • 2012年9月1日以降の各メンバーの最初の予約をリストします
  • 各行に合計メンバーカウントを含むメンバー名のリストを作成します
  • メンバーの番号付きリストを作成します
  • 予約されたスロットの数が最も多い施設IDを出力します。
  • 使用される（丸い）時間のランクメンバー
  • 上位3つの収益生成施設を見つけてください
  • 価値によって施設を分類します
  • 各施設の回収時間を計算します
  • 総収益のローリング平均を計算します
• タイムスタンプを使用する
  • 2012年8月31日に午前1時にタイムスタンプを作成する
  • 互いにタイムスタンプを減算します
  • 2012年10月にすべての日付のリストを生成する
  • タイムスタンプから月の日を取得してください
  • タイムスタンプ間で秒数を計算します
  • 2012年の各月の日数を計算する
  • 月に残っている日数を解決する
  • 予約の終わりの時間を計算します
  • 毎月予約のカウントを返します
  • 月ごとに各施設の利用率を計算します
• 文字列操作
  • メンバーの名前をフォーマットします
  • 名前のプレフィックスで機能を見つけます
  • ケースに依存しない検索を実行します
  • 括弧付きの電話番号を見つけます
  • 主要なゼロを備えたパッドジップコード
  • 姓がアルファベットの各文字から始まるメンバーの数を数える
  • 電話番号をクリーンアップします
• 再帰クエリ
  • メンバーID 27の上向きの推奨チェーンを見つけます
  • メンバーID 1の下向きの推奨チェーンを見つけます
  • 任意のメンバーの上向きの推奨チェーンを返すことができるCTEを作成する

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はじめる

エクササイズを行うのは非常に簡単です。あなたがしなければならないのは、エクササイズを開いて、質問を見て、答えようとすることだけです！

これらの演習のデータセットは、新しく作成されたカントリークラブ向けで、メンバー、テニスコートなどの施設、およびそれらの施設の予約履歴があります。とりわけ、クラブは情報を使用して施設の使用/需要を分析する方法を理解したいと考えています。注：このデータセットは、興味深い一連のエクササイズをサポートするために純粋に設計されており、データベーススキーマにはいくつかの面で欠陥があります。優れたデザインの例として取得しないでください。メンバーのテーブルを見てみましょう。

 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
);

各メンバーには、ID（シーケンシャルであることは保証されていません）、基本的なアドレス情報、それらを推奨するメンバーへの参照（もしあれば）、およびそれらが参加したときのタイムスタンプがあります。データセットのアドレスは、完全に（そして非現実的に）製造されています。

 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)
);

施設のテーブルには、カントリークラブが持っているすべての予約可能な施設がリストされています。クラブには、ID/名前情報、メンバーとゲストの両方の予約コスト、施設の建設の初期コスト、および推定毎月の維持費が保存されています。彼らは、この情報を使用して、各施設がどれほど財政的に価値があるかを追跡したいと考えています。

 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)
);

最後に、施設の予約を追跡するテーブルがあります。これにより、施設ID、予約を行ったメンバー、予約の開始、予約が行われた30分の「スロット」があります。この特異なデザインは、特定のクエリをより困難にしますが、いくつかの興味深い課題を提供する必要があります。

さて、それはあなたが必要とするすべての情報でなければなりません。上記のメニューから試すために、クエリのカテゴリを選択するか、または最初から開始することもできます。

自分のPostgresシステムを使用したいです

問題ない！起きて実行することはそれほど難しくありません。まず、PostgreSQLのインストールが必要です。これをここから入手できます。開始したら、SQLをダウンロードします。

最後に、 psql -U <username> -f clubdata.sql -d postgres -x -qを実行して「エクササイズ」データベースを作成します。ポストグレス「pgexercise」ユーザー、テーブル、データをロードします。 Cロケールを使用します）

クエリを実行しているときは、PSQLが少し不格好だと感じるかもしれません。その場合、PGADMINまたはEclipseデータベース開発ツールを試すことをお勧めします。

スキーマ

---

簡単なSQLクエリ

このカテゴリは、SQLの基本を扱います。それは、選択と条項、ケース表現、組合、および他のいくつかのオッズと終わりをカバーします。すでにSQLで教育を受けている場合は、おそらくこれらの演習がかなり簡単だと思うでしょう。そうでない場合は、今後のより困難なカテゴリの学習を開始するための適切なポイントを見つける必要があります！

これらの質問に苦労している場合は、Alan BeaulieuのSQLを、このテーマに関する簡潔でよく書かれた本として学習することを強くお勧めします。データベースシステムの基礎に興味がある場合（それらの使用方法とは対照的に）、CJ日付までのデータベースシステムの紹介も調査する必要があります。

テーブルからすべてを取得します

CD.Facitiesテーブルからすべての情報をどのように取得できますか？

期待される結果：

facid	名前	メンバーコスト	GuestCost	InitialOutlay	毎月のメンテナンス
0	テニスコート1	5	25	10000	200
1	テニスコート2	5	25	8000	200
2	バドミントン裁判所	0	15.5	4000	50
3	卓球	0	5	320	10
4	マッサージルーム1	35	80	4000	3000
5	マッサージルーム2	35	80	4000	3000
6	スカッシュコート	3.5	17.5	5000	80
7	スヌーカーテーブル	0	5	450	15
8	ビリヤード台	0	5	400	15

答え：

 select * from cd . facilities ;          

SELECTステートメントは、データベースから情報を読み取るクエリの基本的な開始ブロックです。通常、最小選択ステートメントはselect [some set of columns] from [some table or group of tables]で構成されています。

この場合、施設テーブルからすべての情報が必要です。 From Sectionは簡単です - cd.facilitiesテーブルを指定するだけです。 「CD」はテーブルのスキーマです。これは、データベース内の関連情報の論理的なグループ化に使用される用語です。

次に、すべての列が必要であることを指定する必要があります。便利なことに、「すべての列」 - *の速記があります。すべての列名を骨の折れる代わりに、これを使用できます。

テーブルから特定の列を取得します

すべての施設のリストとその費用をメンバーに印刷したいと思います。施設の名前とコストのみのリストをどのように取得しますか？

期待される結果：

名前	メンバーコスト
テニスコート1	5
テニスコート2	5
バドミントン裁判所	0
卓球	0
マッサージルーム1	35
マッサージルーム2	35
スカッシュコート	3.5
スヌーカーテーブル	0
ビリヤード台	0

答え：

 select name, membercost from cd . facilities ;          

この質問では、必要な列を指定する必要があります。 SELECTステートメントに指定された列名の単純なコンマ区切りリストでそれを行うことができます。以下に示すように、すべてのデータベースが句で使用できる列を見ることと、要求した列を返します。

一般的に言えば、非投影クエリの場合、 *を使用するのではなく、クエリに必要な列の名前を指定することが望ましいと考えられています。これは、より多くの列がテーブルに追加された場合、アプリケーションが対処できない可能性があるためです。

どの行が取得されるかを制御します

メンバーに料金を請求する施設のリストをどのように作成できますか？

期待される結果：

facid	名前	メンバーコスト	GuestCost	InitialOutlay	毎月のメンテナンス
0	テニスコート1	5	25	10000	200
1	テニスコート2	5	25	8000	200
4	マッサージルーム1	35	80	4000	3000
5	マッサージルーム2	35	80	4000	3000
6	スカッシュコート	3.5	17.5	5000	80

答え：

 select * from cd . facilities where membercost > 0 ;          

FROM句は、結果を読むために一連の候補行を構築するために使用されます。これまでの例では、この行のセットは単にテーブルの内容でした。将来的には、より興味深い候補者を作成できるようにする参加を探索します。

候補の行のセットを作成したら、 WHERE句を使用すると、興味のある行、この場合はメンバーコストがゼロ以上の行をフィルタリングできます。後の演習でわかるように、 WHEREにはブールロジックと複数のコンポーネントが組み合わされていることがあります。たとえば、0を超えるコストと10未満のコストで施設WHERE検索することが可能です。

どの行を取得するかを制御、パート2

メンバーに料金を請求する施設のリストをどのように作成できますか？その料金は毎月のメンテナンスコストの1/50未満ですか？問題の施設のFACID、施設名、会員費用、および毎月のメンテナンスを返却してください。

期待される結果：

facid	名前	メンバーコスト	毎月のメンテナンス
4	マッサージルーム1	35	3000
5	マッサージルーム2	35	3000

答え：

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

WHEREは、私たちが興味を持っている行のフィルタリングを可能にすることができます - この場合、メンバーコストがゼロを超え、毎月のメンテナンスコストの1/50未満の人。ご覧のとおり、マッサージルームは人材派遣コストのおかげで非常に高価です！

2つ以上の条件をテストしたい場合は、それらを使用しAND結合します。予想通り、使用するORテストすることができます。

これが、 WHERE句と特定の列の選択を組み合わせた最初のクエリであることに気付いたかもしれません。これの効果の下の画像で見ることができます。選択した列と選択した行の交差点により、返されるデータが得られます。これは今ではあまり面白くないかもしれませんが、後で結合するようなより複雑な操作を追加すると、この動作の単純な優雅さがわかります。

基本的な文字列検索

「テニス」という言葉を名前にして、すべての施設のリストをどのように作成できますか？

期待される結果：

facid	名前	メンバーコスト	GuestCost	InitialOutlay	毎月のメンテナンス
0	テニスコート1	5	25	10000	200
1	テニスコート2	5	25	8000	200
3	卓球	0	5	320	10

答え：

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

SQLのLIKEオペレーターは、文字列にシンプルなパターンマッチングを提供します。それは非常に普遍的に実装されており、使いやすく、簡単に使用できます。任意の文字列に一致し、_単一の文字に一致する_を持つ文字列が必要です。この場合、「テニス」という単語を含む名前を探しているので、どちらの側にも％を置くことは法案に適合します。

このタスクを達成する他の方法があります。たとえば、Postgresは〜演算子との正規表現をサポートしています。快適に感じるものは何でも使用しますが、システム間ではるかにポータブルであるLIKE演算子がはるかにポータブルであることに注意してください。

複数の可能な値と一致します

ID 1と5の施設の詳細をどのように取得できますか？またはオペレーターを使用せずに行うようにしてください。

期待される結果：

facid	名前	メンバーコスト	GuestCost	InitialOutlay	毎月のメンテナンス
1	テニスコート2	5	25	8000	200
5	マッサージルーム2	35	80	4000	3000

答え：

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

この質問に対する明らかな答えはwhere facid = 1 or facid = 5ように見えるWHEREを使用することです。多数の可能性のある一致で簡単な代替案は、 INです。 INオペレーターは、可能な値のリストを取得し、（この場合）FACIDと一致します。値の1つが一致する場合、句はその行に対して真であり、行が返されます。

IN Operatorは、リレーショナルモデルの優雅さの優れた初期のデモンストレーターです。それが取る議論は、単なる値のリストではありません - 実際には単一の列を持つテーブルです。クエリもテーブルを返すため、単一の列を返すクエリを作成する場合、それらの結果をINに送ることができます。おもちゃを与えるために：

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

この例は、すべての施設を選択するだけで機能的に同等ですが、あるクエリの結果を別のクエリに供給する方法を示します。内部クエリはサブクエリと呼ばれます。

結果をバケットに分類します

毎月のメンテナンスコストが100ドルを超えるかどうかに応じて、それぞれが「安い」または「高価」とラベル付けされている施設のリストをどのように作成できますか？問題の施設の名前と毎月のメンテナンスを返します。

期待される結果：

名前	料金
テニスコート1	高い
テニスコート2	高い
バドミントン裁判所	安い
卓球	安い
マッサージルーム1	高い
マッサージルーム2	高い
スカッシュコート	安い
スヌーカーテーブル	安い
ビリヤード台	安い

答え：

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

この演習には、いくつかの新しい概念が含まれています。 1つ目は、 SELECTとFROM間のクエリの領域で計算を行っているという事実です。以前は、これを使用して返品したい列のみを選択していましたが、サブクエリを含む、返された行ごとに単一の結果を生成するものをここに置くことができます。

2番目の新しい概念は、 CASEステートメント自体です。 CASE 、クエリに示されているフォームを使用して、他の言語のif/switchステートメントと効果的に似ています。 「中間」オプションを追加するために、いつ別のwhen...thenを挿入します。

最後に、 ASオペレーターがあります。これは、列または表現のラベルを付けるために単に使用され、それらをより適切に表示するか、サブクエリの一部として使用すると参照しやすくするために使用されます。

日付を扱う

2012年9月の開始後に参加したメンバーのリストをどのように作成できますか？問題のメンバーのmemid、surname、firstName、およびJoindateを返します。

期待される結果：

memid	姓	ファーストネーム	ジョインド
24	サーウィン	Ramnaresh	2012-09-01 08:44:42
26	ジョーンズ	ダグラス	2012-09-02 18:43:05
27	ラミー	ヘンリエッタ	2012-09-05 08:42:35
28	ファレル	デビッド	2012-09-15 08:22:05
29	ワージントンスミス	ヘンリー	2012-09-17 12:27:15
30	purview	ミリセント	2012-09-18 19:04:01
33	タッパーウェア	ヒヤシンス	2012-09-18 19:32:05
35	ハント	ジョン	2012-09-19 11:32:45
36	クランペット	エリカ	2012-09-22 08:36:38
37	スミス	ダレン	2012-09-26 18:08:45

答え：

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

これは、SQLタイムスタンプの最初の外観です。それらは、下降順にフォーマットされています： YYYY-MM-DD HH:MM:SS.nnnnnn 。日付間の違いを得ることはもう少し関与していますが（そして強力です！）、Unixタイムスタンプと同じように比較できます。この場合、タイムスタンプの日付部分を指定しました。これは、Postgresによって完全なタイムスタンプ2012-09-01 00:00:00に自動的にキャストされます。

重複を削除し、結果を注文します

メンバーテーブルに最初の10姓の注文されたリストをどのように作成できますか？リストには複製を含めてはなりません。

期待される結果：

姓
より悪い
ベイカー
ブース
バター
コプリン
クランペット
あえて
ファレル
ゲスト
紳士

答え：

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

ここには3つの新しい概念がありますが、それらはすべて非常に簡単です。

• SELECT後にDISTINCT指定は、結果セットから重複する行を削除します。これは行に適用されることに注意してください。行Aに複数の列がある場合、行Bはすべての列の値が同じ場合にのみ等しくなります。一般的なルールとして、意志のあるファッションでDISTINCT使用を使用しないでください - 大規模なクエリ結果セットから重複を削除することは自由ではないので、必要に応じて行います。
• （クエリの終わり近くの句の後とFROMの後とWHERE ） ORDER BYを指定すると、列または列のセット（コンマ分離）で結果を順序付けることができます。
• LIMITキーワードを使用すると、取得した結果の数を制限できます。これは、結果のページを一度に入手するのに役立ち、 OFFSETキーワードと組み合わせて次のページを取得できます。これはMySQLで使用されているのと同じアプローチであり、非常に便利です - 残念ながら、このプロセスは他のDBSではもう少し複雑であることがわかります。

複数のクエリからの結果を組み合わせます

何らかの理由で、すべての姓とすべての施設名の複合リストが必要です。はい、これは不自然な例です:-)。そのリストを作成します！

期待される結果：

姓
テニスコート2
ワージントンスミス
バドミントン裁判所
ピンカー
あえて
より悪い
マッケンジー
クランペット
マッサージルーム1
スカッシュコート

答え：

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

UNIONオペレーターは、予想されることを行います。2つのSQLクエリの結果を1つのテーブルに組み合わせます。警告は、2つのクエリの両方の結果の両方が同じ数の列と互換性のあるデータ型を持っている必要があることです。

UNION重複する行を削除しますが、 UNION ALLそうではありません。重複した結果を気にしない限り、デフォルトでUNION ALLデフォルトで使用します。

単純な集約

最後のメンバーのサインアップ日を取得したいと思います。この情報をどのように取得できますか？

期待される結果：

最新
2012-09-26 18:08:45

答え：

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

これは、SQLの集計関数への最初の進出です。それらは、行のグループ全体に関する情報を抽出し、次のような質問を簡単に尋ねるために使用されます。

• 毎月維持するのに最も高価な施設は何ですか？
• 誰が最も新しいメンバーを推薦しましたか？
• 各メンバーは私たちの施設でどのくらいの時間を費やしましたか？

ここでの最大集約関数は非常に単純です。Joindateのすべての可能な値を受信し、最大の値を出力します。関数を集約するためのより多くの力がありますが、将来の演習で出会うことになります。

より多くの集約

日付だけでなく、サインアップした最後のメンバーの最初と姓を取得したいと思います。どうすればできますか？

期待される結果：

ファーストネーム	姓	ジョインド
ダレン	スミス	2012-09-26 18:08:45

答え：

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

上記の提案されたアプローチでは、サブクエリを使用して、最新のJoindateが何であるかを調べます。このサブクエリは、スカラーテーブル、つまり、単一の列と単一の行を備えたテーブルを返します。単一の値しかないので、単一の一定の値を置く可能性のある場所にサブクエリを置き換えることができます。この場合、それを使用して、特定のメンバーを見つけるためにWHEREの句句を完成させます。

あなたは以下のようなことができることを願っています：

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

残念ながら、これはうまくいきません。 MAX関数は、 WHEREのような行を制限しません - それは単に多くの価値を取り入れ、最大の値を返します。次に、データベースは、名前の長いリストを最大関数から出てくる単一の参加日をペアリングする方法を疑問にしておきます。代わりに、「最大参加日と同じ参加日がある行を見つけてください」と言わなければなりません。

ヒントが述べたように、この仕事を成し遂げる他の方法があります - 一例は以下にあります。このアプローチでは、最後の参加日が何であるかを明示的に見つけるのではなく、参加日を下ってメンバーテーブルを注文し、最初のテーブルを選択します。このアプローチは、まったく同時に参加する2人の非常にありそうもない不測の事態をカバーしていないことに注意してください:-)。

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

---

参加してサブクリーリー

このカテゴリは、主にリレーショナルデータベースシステムの基本的な概念を扱います。結合。結合することで、複数のテーブルから関連情報を組み合わせて質問に答えることができます。これは、クエリの容易さに有益であるだけではありません。結合能力の欠如は、データの非規制を促進し、データを内部的に一貫性を保つ複雑さを高めます。

このトピックは、内側、外側、および自己結合をカバーし、サブクリーリー（クエリ内のクエリ）に少し時間を費やします。これらの質問に苦労している場合は、Alan BeaulieuのSQLを、このテーマに関する簡潔でよく書かれた本として学習することを強くお勧めします。

メンバーの予約の開始時間を取得します

「David Farrell」という名前のメンバーによる予約の開始時間のリストをどのように作成できますか？

期待される結果：

開始時刻
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

答え：

 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 ' ;          

最も一般的に使用される一種の結合はINNER JOINです。これが行うことは、結合式に基づいて2つのテーブルを組み合わせることです。この場合、メンバーテーブルの各メンバーIDについて、予約テーブルの一致する値を探しています。一致が見つかる場所では、各テーブルの値を組み合わせた行が返されます。各テーブルにエイリアス（BKSとMEMS）を与えたことに注意してください。これは2つの理由で使用されます。1つ目は便利であり、次に同じテーブルに何度か参加し、テーブルが結合された各時間と列を区別する必要があります。

私たちの選択と今のところ条項を無視し、 FROM Statementが生み出しているものに焦点を当てましょう。以前のすべての例では、 FROM単純なテーブルでした。今は何ですか？別のテーブル！今回は、予約とメンバーの複合として制作されています。以下の結合の出力のサブセットを見ることができます。

メンバーテーブルの各メンバーについて、Joinは予約テーブルのすべてのマッチングメンバーIDを見つけました。その後、各試合で、メンバーテーブルの行と予約テーブルから行を組み合わせた行を作成しました。

明らかに、これはそれ自体があまりにも多くの情報であり、有用な質問はそれを除外したいと思うでしょう。クエリでは、以下に示すように、 SELECT句の開始を使用して列を選択し、列を選択してWHEREを選択します。

デビッドの予約を見つけるために必要なのはそれだけです！一般的に、 FROM句の出力は本質的に1つの大きなテーブルであり、その後情報をフィルタリングすることを覚えておくことをお勧めします。これは非効率的に聞こえるかもしれませんが、心配しないでください、カバーの下では、DBはよりインテリジェントに動作します:-)。

最後の注：内側結合には2つの異なる構文があります。私が好むものをあなたに示しました、私は他の結合タイプとより一致していると思うことを示しました。一般に、以下に示す別の構文が表示されます。

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

これは、機能的には承認された答えとまったく同じです。この構文をより快適に感じる場合は、お気軽に使用してください！

テニスコートの予約の開始時間を実現する

日付「2012-09-21」のテニスコートの予約の開始時間のリストをどのように作成できますか？開始時間と施設名のペアリングのリストを返し、時間ごとに注文します。

期待される結果：

始める	名前
2012-09-21 08:00:00	テニスコート1
2012-09-21 08:00:00	テニスコート2
2012-09-21 09:30:00	テニスコート1
2012-09-21 10:00:00	テニスコート2
2012-09-21 11:30:00	テニスコート2
2012-09-21 12:00:00	テニスコート1
2012-09-21 13:30:00	テニスコート1
2012-09-21 14:00:00	テニスコート2
2012-09-21 15:30:00	テニスコート1
2012-09-21 16:00:00	テニスコート2
2012-09-21 17:00:00	テニスコート1
2012-09-21 18:00:00	テニスコート2

答え：

 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 ;          

これは別のINNER JOINクエリですが、それはかなり複雑です！クエリのFROMから簡単です - 私たちは単にFACIDで施設と予約のテーブルに一緒に参加しています。これにより、予約の各行について、予約されている施設に関する詳細な情報を添付したテーブルが生成されます。

WHEREのコンポーネントに。開始時刻のチェックはかなり自明です - すべての予約が指定された日付間で開始されることを確認しています。私たちはテニスコートのみに関心があるため、 INオペレーターを使用して、データベースシステムに施設ID 0または1のみを返すように指示します - 裁判所のIDです。これを表現する他の方法があります： where facs.facid = 0 or facs.facid = 1 、またはwhere facs.name like 'Tennis%'でも、facs.facid = 0またはfacs.facid = 1を使用できたのです。

残りは非常にシンプルです。興味のある列SELECT 、開始時間ORDER BY 。

別のメンバーを推薦したすべてのメンバーのリストを作成する

別のメンバーを推薦したすべてのメンバーのリストをどのように出力できますか？リストに重複がないことを確認し、その結果が（姓、FirstName）によって注文されていることを確認してください。

期待される結果：

ファーストネーム	姓
フィレンツェ	より悪い
ティモシー	ベイカー
ジェラルド	バター
ジェミマ	ファレル
マシュー	紳士
デビッド	ジョーンズ
ジャニス	ジョプレット
ミリセント	purview
ティム	Rownam
ダレン	スミス
トレーシー	スミス
熟考	スティボン
バートン	トレーシー

答え：

 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;          

一部の人々が混乱を感じる概念は次のとおりです。あなたはそれ自体にテーブルに参加できます！これは、Cd.membersの推奨されるものと同様に、同じ表にデータを参照する列がある場合に非常に便利です。

これを視覚化するのに苦労している場合は、これが他の内側の結合と同じように機能することを忘れないでください。私たちの参加は、推奨される値を持つメンバーの各行を取り、一致するメンバーIDを持つ行を再びメンバーを探します。次に、2つのメンバーエントリを組み合わせた出力行を生成します。これは以下の図のように見えます：

出力セットに2つの「姓」列があるかもしれませんが、テーブルエイリアスで区別できることに注意してください。希望する列を選択したら、 DISTINCTを使用して、重複がないことを確認します。

すべてのメンバーのリストを作成し、推奨

推奨する個人を含むすべてのメンバーのリストをどのように出力できますか？結果が（姓、firstName）で注文されることを確認してください。

期待される結果：

memfname	memsname	recfname	recsname
フィレンツェ	より悪い	熟考	スティボン
アン	ベイカー	熟考	スティボン
ティモシー	ベイカー	ジェミマ	ファレル
ティム	ブース	ティム	Rownam
ジェラルド	バター	ダレン	スミス
ジョーン	コプリン	ティモシー	ベイカー
エリカ	クランペット	トレーシー	スミス
ナンシー	あえて	ジャニス	ジョプレット
デビッド	ファレル
ジェミマ	ファレル
ゲスト	ゲスト
マシュー	紳士	ジェラルド	バター
ジョン	ハント	ミリセント	purview
デビッド	ジョーンズ	ジャニス	ジョプレット
ダグラス	ジョーンズ	デビッド	ジョーンズ
ジャニス	ジョプレット	ダレン	スミス
アンナ	マッケンジー	ダレン	スミス
チャールズ	オーウェン	ダレン	スミス
デビッド	ピンカー	ジェミマ	ファレル
ミリセント	purview	トレーシー	スミス
ティム	Rownam
ヘンリエッタ	ラミー	マシュー	紳士
Ramnaresh	サーウィン	フィレンツェ	より悪い
ダレン	スミス
ダレン	スミス
ジャック	スミス	ダレン	スミス
トレーシー	スミス
熟考	スティボン	バートン	トレーシー
バートン	トレーシー
ヒヤシンス	タッパーウェア
ヘンリー	ワージントンスミス	トレーシー	スミス

答え：

 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;          

別の新しい概念を紹介しましょう： LEFT OUTER JOIN 。これらは、内側の結合とは異なる方法によって最もよく説明されています。内側の結合は、左と右のテーブルを取り、結合条件（ ON ）に基づいて一致する行を探します。条件が満たされると、結合された行が生成されます。 LEFT OUTER JOIN同様に動作しますが、左手テーブルの特定の行が何も一致しない場合でも、出力行が生成されます。その出力行は、左手のテーブルの列と、右手のテーブル列の代わりにたくさんのNULLSで構成されています。

これは、この質問のような状況で役立ちます。この質問では、オプションのデータを使用して出力を生成したいと考えています。私たちはすべてのメンバーの名前と、その人が存在する場合は彼らの推薦者の名前を望んでいます。内側の結合でそれを適切に表現することはできません。

ご想像のとおり、他の外側の結合もあります。 RIGHT OUTER JOIN 、式の左側がオプションのデータを含むものであることを除いて、 LEFT OUTER JOINによく似ています。まれに使用されていないFULL OUTER JOIN式の両側をオプションとして扱います。

テニスコートを使用したすべてのメンバーのリストを作成する

テニスコートを使用したすべてのメンバーのリストをどのように作成できますか？あなたの出力に裁判所の名前を含め、メンバーの名前は単一の列としてフォーマットされています。複製データがないことを確認し、メンバー名で注文します。

期待される結果：

メンバー	施設
アン・ベイカー	テニスコート2
アン・ベイカー	テニスコート1
バートン・トレーシー	テニスコート2
バートン・トレーシー	テニスコート1
チャールズオーウェン	テニスコート2
チャールズオーウェン	テニスコート1
ダレン・スミス	テニスコート2
デビッド・ファレル	テニスコート2
デビッド・ファレル	テニスコート1
デビッド・ジョーンズ	テニスコート1
デビッド・ジョーンズ	テニスコート2
デビッド・ピンカー	テニスコート1
ダグラス・ジョーンズ	テニスコート1
エリカ・クランペット	テニスコート1
フローレンス・バダー	テニスコート1
フローレンス・バダー	テニスコート2
ゲスト	テニスコート2
ゲスト	テニスコート1
ジェラルドバター	テニスコート1
ジェラルドバター	テニスコート2
ヘンリエッタ・ラニー	テニスコート2
ジャック・スミス	テニスコート1
ジャック・スミス	テニスコート2
ジャニス・ジョプレット	テニスコート1
ジャニス・ジョプレット	テニスコート2
ジェミマ・ファレル	テニスコート2
ジェミマ・ファレル	テニスコート1
ジョーン・コプリン	テニスコート1
ジョン・ハント	テニスコート1
ジョン・ハント	テニスコート2
マシュー・ゲンティン	テニスコート1
ミリセントパビュー	テニスコート2
ナンシー・デア	テニスコート2
ナンシー・デア	テニスコート1
熟考するスティボン	テニスコート2
熟考するスティボン	テニスコート1
Ramnaresh Sarwin	テニスコート2
Ramnaresh Sarwin	テニスコート1
ティムブース	テニスコート1
ティムブース	テニスコート2
ティム・ロウネム	テニスコート1
ティム・ロウネム	テニスコート2
ティモシー・ベイカー	テニスコート2
ティモシー・ベイカー	テニスコート1
トレーシー・スミス	テニスコート2
トレーシー・スミス	テニスコート1

答え：

 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          

この演習は、主に以前の質問で学んだことのより複雑なアプリケーションです。また、複数の参加を使用したのは初めてです。これは、一部の人にとっては少し混乱するかもしれません。結合式を読むとき、結合は効果的に2つのテーブルを取る関数であり、1つは左のテーブルにラベル付けされ、もう1つは右側にラベルを付けることに注意してください。これは、クエリに1つだけ結合するだけで視覚化するのが簡単ですが、2つとはもう少し混乱しています。

このクエリでの2番目INNER JOIN cd.facitiesの右側があります。それは把握するのに十分簡単です。ただし、左側は、Cd.membersをCd.Bookingsに結合することで返されたテーブルです。これを強調することが重要です。リレーショナルモデルはすべてテーブルに関するものです。任意の結合の出力は別のテーブルです。クエリの出力はテーブルです。単一の列リストはテーブルです。それを把握すると、モデルの基本的な美しさを把握しました。

最後のメモとして、ここに1つの新しいものを紹介します： ||オペレーターは、文字列を連結するために使用されます。

高価な予約のリストを作成します

2012-09-14の日に予約のリストを作成するには、メンバー（またはゲスト）が30ドルを超える費用がかかりますか？ゲストはメンバーに異なるコストを持っていることを忘れないでください（リストされているコストは30分あたり「スロット」です）。ゲストユーザーは常にID 0です。出力には、施設の名前、メンバーの名前が単一の列としてフォーマットされたものとコストを含めます。下降コストで注文し、サブクリーリーを使用しないでください。

期待される結果：

メンバー	施設	料金
ゲスト	マッサージルーム2	320
ゲスト	マッサージルーム1	160
ゲスト	マッサージルーム1	160
ゲスト	マッサージルーム1	160
ゲスト	テニスコート2	150
ジェミマ・ファレル	マッサージルーム1	140
ゲスト	テニスコート1	75
ゲスト	テニスコート2	75
ゲスト	テニスコート1	75
マシュー・ゲンティン	マッサージルーム1	70
フローレンス・バダー	マッサージルーム2	70
ゲスト	スカッシュコート	70.0
ジェミマ・ファレル	マッサージルーム1	70
熟考するスティボン	マッサージルーム1	70
バートン・トレーシー	マッサージルーム1	70
ジャック・スミス	マッサージルーム1	70
ゲスト	スカッシュコート	35.0
ゲスト	スカッシュコート	35.0

答え：

 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 ;          

これは少し複雑なものです！以前に使用したよりも複雑なロジックですが、発言することはあまりありません。 WHERE句は、2012-09-14の出力を2012-09-14に十分に費用のかかる列に制限し、ゲストと他の人を区別することを忘れないでください。次に、列選択のCASEステートメントを使用して、メンバーまたはゲストの正しいコストを出力します。

すべてのメンバーのリストを作成し、参加者を使用しないでください

参加者を使用せずに（もしあれば）個人を含むすべてのメンバーのリストをどのように出力できますか？リストに重複がないこと、および各firstName + surnameペアリングが列としてフォーマットされ、注文されることを確認してください。

期待される結果：

メンバー	推奨
アンナ・マッケンジー	ダレン・スミス
アン・ベイカー	熟考するスティボン
バートン・トレーシー
チャールズオーウェン	ダレン・スミス
ダレン・スミス
デビッド・ファレル
デビッド・ジョーンズ	ジャニス・ジョプレット
デビッド・ピンカー	ジェミマ・ファレル
ダグラス・ジョーンズ	デビッド・ジョーンズ
エリカ・クランペット	トレーシー・スミス
フローレンス・バダー	熟考するスティボン
ゲスト
ジェラルドバター	ダレン・スミス
ヘンリエッタ・ラニー	マシュー・ゲンティン
ヘンリー・ワージントン・スミス	トレーシー・スミス
Hyacinth Tupperware
ジャック・スミス	ダレン・スミス
ジャニス・ジョプレット	ダレン・スミス
ジェミマ・ファレル
ジョーン・コプリン	ティモシー・ベイカー
ジョン・ハント	ミリセントパビュー
マシュー・ゲンティン	ジェラルドバター
ミリセントパビュー	トレーシー・スミス
ナンシー・デア	ジャニス・ジョプレット
Ponder Stibbons	Burton Tracy
Ramnaresh Sarwin	Florence Bader
Tim Boothe	Tim Rownam
Tim Rownam
Timothy Baker	Jemima Farrell
Tracy Smith

答え：

 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:

メンバー	施設	料金
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
ジャック・スミス	Massage Room 1	70
GUEST GUEST	Squash Court	35.0
GUEST GUEST	Squash Court	35.0

答え：

 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 .

---

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	名前	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	卓球	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	スパ	20	30	100000	800

答え：

 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	名前	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	卓球	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	スパ	20	30	100000	800
10	Squash Court 2	3.5	17.5	5000	80

答え：

 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 .例えば：

 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	名前	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	卓球	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	スパ	20	30	100000	800

答え：

 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	名前	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	卓球	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

答え：

 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	名前	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	卓球	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

答え：

 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	名前	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	卓球	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

答え：

 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

答え：

 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	姓	ファーストネーム	住所	郵便番号	電話	recommendedby	joindate
0	ゲスト	ゲスト	ゲスト	0	[phone]		2012-07-01 00:00:00
1	Smith	ダレン	8 Bloomsbury Close, Boston	4321	555-555-5555		2012-07-02 12:02:05
2	Smith	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	ジャニス	20 Crossing Road, New York	234	[phone]	1	2012-07-03 10:25:05
5	Butters	ジェラルド	1065 Huntingdon Avenue, Boston	56754	[phone]	1	2012-07-09 10:44:09
6	Tracy	バートン	3 Tunisia Drive, Boston	45678	[phone]		2012-07-15 08:52:55
7	Dare	ナンシー	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	オーウェン	チャールズ	52 Cheshire Grove, Winchester, 28563	28563	[phone]	1	2012-08-03 19:42:37
11	ジョーンズ	デビッド	976 Gnats Close, Reading	33862	[phone]	4	2012-08-06 16:32:55
12	ベイカー	アン	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	Smith	ジャック	252 Binkington Way, Boston	69302	[phone]	1	2012-08-10 16:22:05
15	Bader	フィレンツェ	264 Ursula Drive, Westford	84923	[phone]	9	2012-08-10 17:52:03
16	ベイカー	ティモシー	329 James Street, Reading	58393	833-941-0824	13	2012-08-15 10:34:25
17	Pinker	デビッド	5 Impreza Road, Boston	65332	811 409-6734	13	2012-08-16 11:32:47
20	Genting	マシュー	4 Nunnington Place, Wingfield, Boston	52365	[phone]	5	2012-08-19 14:55:55
21	Mackenzie	アンナ	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	ジョーンズ	ダグラス	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	デビッド	437 Granite Farm Road, Westford	43532	[phone]		2012-09-15 08:22:05
29	Worthington-Smyth	ヘンリー	55 Jagbi Way, North Reading	97676	[phone]	2	2012-09-17 12:27:15
30	Purview	Millicent	641 Drudgery Close, Burnington, Boston	34232	[phone]	2	2012-09-18 19:04:01
33	Tupperware	Hyacinth	33 Cheerful Plaza, Drake Road, Westford	68666	[phone]		2012-09-18 19:32:05
35	ハント	ジョン	5 Bullington Lane, Boston	54333	[phone]	30	2012-09-19 11:32:45
36	Crumpet	エリカ	Crimson Road, North Reading	75655	[phone]	2	2012-09-22 08:36:38

答え：

 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	姓	ファーストネーム	住所	郵便番号	電話	recommendedby	joindate
0	ゲスト	ゲスト	ゲスト	0	[phone]		2012-07-01 00:00:00
1	Smith	ダレン	8 Bloomsbury Close, Boston	4321	555-555-5555		2012-07-02 12:02:05
2	Smith	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	ジャニス	20 Crossing Road, New York	234	[phone]	1	2012-07-03 10:25:05
5	Butters	ジェラルド	1065 Huntingdon Avenue, Boston	56754	[phone]	1	2012-07-09 10:44:09
6	Tracy	バートン	3 Tunisia Drive, Boston	45678	[phone]		2012-07-15 08:52:55
7	Dare	ナンシー	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	オーウェン	チャールズ	52 Cheshire Grove, Winchester, 28563	28563	[phone]	1	2012-08-03 19:42:37
11	ジョーンズ	デビッド	976 Gnats Close, Reading	33862	[phone]	4	2012-08-06 16:32:55
12	ベイカー	アン	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	Smith	ジャック	252 Binkington Way, Boston	69302	[phone]	1	2012-08-10 16:22:05
15	Bader	フィレンツェ	264 Ursula Drive, Westford	84923	[phone]	9	2012-08-10 17:52:03
16	ベイカー	ティモシー	329 James Street, Reading	58393	833-941-0824	13	2012-08-15 10:34:25
17	Pinker	デビッド	5 Impreza Road, Boston	65332	811 409-6734	13	2012-08-16 11:32:47
20	Genting	マシュー	4 Nunnington Place, Wingfield, Boston	52365	[phone]	5	2012-08-19 14:55:55
21	Mackenzie	アンナ	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	ジョーンズ	ダグラス	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	デビッド	437 Granite Farm Road, Westford	43532	[phone]		2012-09-15 08:22:05
29	Worthington-Smyth	ヘンリー	55 Jagbi Way, North Reading	97676	[phone]	2	2012-09-17 12:27:15
30	Purview	Millicent	641 Drudgery Close, Burnington, Boston	34232	[phone]	2	2012-09-18 19:04:01
33	Tupperware	Hyacinth	33 Cheerful Plaza, Drake Road, Westford	68666	[phone]		2012-09-18 19:32:05
35	ハント	ジョン	5 Bullington Lane, Boston	54333	[phone]	30	2012-09-19 11:32:45
36	Crumpet	エリカ	Crimson Road, North Reading	75655	[phone]	2	2012-09-22 08:36:38

答え：

 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:

カウント
9

答え：

 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.

カウント
6

答え：

 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:

recommendedby	カウント
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

答え：

 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

答え：

 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

答え：

 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	月	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

答え：

 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:

カウント
30

答え：

 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

答え：

 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:

名前	収益
卓球	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

答え：

 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:

名前	収益
卓球	180
Snooker Table	240
Pool Table	270

答え：

 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

答え：

 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.もっとあります。 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	月	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

答え：

 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	名前	Total Hours
0	Tennis Court 1	660.00
1	Tennis Court 2	639.00
2	Badminton Court	604.50
3	卓球	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

答え：

 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:

姓	ファーストネーム	memid	starttime
ゲスト	ゲスト	0	2012-09-01 08:00:00
Smith	ダレン	1	2012-09-01 09:00:00
Smith	Tracy	2	2012-09-01 11:30:00
Rownam	Tim	3	2012-09-01 16:00:00
Joplette	ジャニス	4	2012-09-01 15:00:00
Butters	ジェラルド	5	2012-09-02 12:30:00
Tracy	バートン	6	2012-09-01 15:00:00
Dare	ナンシー	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
オーウェン	チャールズ	10	2012-09-01 11:00:00
ジョーンズ	デビッド	11	2012-09-01 09:30:00
ベイカー	アン	12	2012-09-01 14:30:00
Farrell	Jemima	13	2012-09-01 09:30:00
Smith	ジャック	14	2012-09-01 11:00:00
Bader	フィレンツェ	15	2012-09-01 10:30:00
ベイカー	ティモシー	16	2012-09-01 15:00:00
Pinker	デビッド	17	2012-09-01 08:30:00
Genting	マシュー	20	2012-09-01 18:00:00
Mackenzie	アンナ	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
ジョーンズ	ダグラス	26	2012-09-08 13:00:00
Rumney	Henrietta	27	2012-09-16 13:30:00
Farrell	デビッド	28	2012-09-18 09:00:00
Worthington-Smyth	ヘンリー	29	2012-09-19 09:30:00
Purview	Millicent	30	2012-09-19 11:30:00
Tupperware	Hyacinth	33	2012-09-20 08:00:00
ハント	ジョン	35	2012-09-23 14:00:00
Crumpet	エリカ	36	2012-09-27 11:30:00

答え：

 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:

カウント	ファーストネーム	姓
31	ゲスト	ゲスト
31	ダレン	Smith
31	Tracy	Smith
31	Tim	Rownam
31	ジャニス	Joplette
31	ジェラルド	Butters
31	バートン	Tracy
31	ナンシー	Dare
31	Tim	Boothe
31	Ponder	Stibbons
31	チャールズ	オーウェン
31	デビッド	ジョーンズ
31	アン	ベイカー
31	Jemima	Farrell
31	ジャック	Smith
31	フィレンツェ	Bader
31	ティモシー	ベイカー
31	デビッド	Pinker
31	マシュー	Genting
31	アンナ	Mackenzie
31	Joan	Coplin
31	Ramnaresh	Sarwin
31	ダグラス	ジョーンズ
31	Henrietta	Rumney
31	デビッド	Farrell
31	ヘンリー	Worthington-Smyth
31	Millicent	Purview
31	Hyacinth	Tupperware
31	ジョン	ハント
31	エリカ	Crumpet
31	ダレン	Smith

答え：

 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	ファーストネーム	姓
1	ゲスト	ゲスト
2	ダレン	Smith
3	Tracy	Smith
4	Tim	Rownam
5	ジャニス	Joplette
6	ジェラルド	Butters
7	バートン	Tracy
8	ナンシー	Dare
9	Tim	Boothe
10	Ponder	Stibbons
11	チャールズ	オーウェン
12	デビッド	ジョーンズ
13	アン	ベイカー
14	Jemima	Farrell
15	ジャック	Smith
16	フィレンツェ	Bader
17	ティモシー	ベイカー
18	デビッド	Pinker
19	マシュー	Genting
20	アンナ	Mackenzie
21	Joan	Coplin
22	Ramnaresh	Sarwin
23	ダグラス	ジョーンズ
24	Henrietta	Rumney
25	デビッド	Farrell
26	ヘンリー	Worthington-Smyth
27	Millicent	Purview
28	Hyacinth	Tupperware
29	ジョン	ハント
30	エリカ	Crumpet
31	ダレン	Smith

答え：

 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	合計
4	1404

答え：

 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:

ファーストネーム	姓	hours	ランク
ゲスト	ゲスト	1200	1
ダレン	Smith	340	2
Tim	Rownam	330	3
Tim	Boothe	220	4
Tracy	Smith	220	4
ジェラルド	Butters	210	6
バートン	Tracy	180	7
チャールズ	オーウェン	170	8
ジャニス	Joplette	160	9
アン	ベイカー	150	10
ティモシー	ベイカー	150	10
デビッド	ジョーンズ	150	10
ナンシー	Dare	130	13
フィレンツェ	Bader	120	14
アンナ	Mackenzie	120	14
Ponder	Stibbons	120	14
ジャック	Smith	110	17
Jemima	Farrell	90	18
デビッド	Pinker	80	19
Ramnaresh	Sarwin	80	19
マシュー	Genting	70	21
Joan	Coplin	50	22
デビッド	Farrell	30	23
ヘンリー	Worthington-Smyth	30	23
ジョン	ハント	20	25
ダグラス	ジョーンズ	20	25
Millicent	Purview	20	25
Henrietta	Rumney	20	25
エリカ	Crumpet	10	29
Hyacinth	Tupperware	10	29

答え：

 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:

名前	ランク
Massage Room 1	1
Massage Room 2	2
Tennis Court 2	3

答え：

 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:

名前	収益
Massage Room 1	高い
Massage Room 2	高い
Tennis Court 2	高い
Badminton Court	平均
Squash Court	平均
Tennis Court 1	平均
Pool Table	低い
Snooker Table	低い
卓球	低い

答え：

 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:

名前	数ヶ月
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
卓球	6.4000000000000000
Tennis Court 1	2.2624434389140271
Tennis Court 2	1.7505470459518600

答え：

 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:

日付	収益
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

答え：

 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

答え：

 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:

間隔
32 days

答え：

 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

答え：

 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

答え：

 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

答え：

 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

答え：

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:

月	長さ
1	31 days
2	29 days
3	31 days
4	30日
5	31 days
6	30日
7	31 days
8	31 days
9	30日
10	31 days
11	30日
12	31 days

答え：

 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:

残り
19 days

答え：

 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

答え：

 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:

月	カウント
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

答え：

 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:

名前	月	利用
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
卓球	2012-07-01 00:00:00	13.4
卓球	2012-08-01 00:00:00	39.2
卓球	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

答え：

 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:

名前
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

答え：

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

答え：

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

答え：

 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	電話
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]

答え：

 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.
• The '.' 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:

ジップ
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

答え：

 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:

手紙	カウント
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

答え：

 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	電話
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

答え：

 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	ファーストネーム	姓
20	マシュー	Genting
5	ジェラルド	Butters
1	ダレン	Smith

答え：

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	ファーストネーム	姓
4	ジャニス	Joplette
5	ジェラルド	Butters
7	ナンシー	Dare
10	チャールズ	オーウェン
11	デビッド	ジョーンズ
14	ジャック	Smith
20	マシュー	Genting
21	アンナ	Mackenzie
26	ダグラス	ジョーンズ
27	Henrietta	Rumney

答え：

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:

メンバー	recommender	ファーストネーム	姓
12	9	Ponder	Stibbons
12	6	バートン	Tracy
22	16	ティモシー	ベイカー
22	13	Jemima	Farrell

答え：

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.