작업에서 발생하는 문제, 두 줄에 일치하는 문제, 요구 사항 : 각 문자열에는 최대 하나가 포함되어 있으며 무한히 다중 일 수 있습니다.
* 길이의 줄을 나타내는 동안? 캐릭터를 나타냅니다
요구 사항은 두 가지 충돌을 유발할 수 있습니다
다음과 같이 코드를 복사하십시오. <input type = "text"id = "str1"> <br>
<입력 유형 = "text"id = "str2"> <br>
<input type = "button"onclick = "checkmarchx ()"value = "check">
JavaScript를 사용하여 구현하려는 코드는 다음과 같습니다.
함수 checkMarchx () {var str1 = document.getElementByid ( 'str1'). str1xposition> str2xposition? str2xposition; // successif (position! = 0) {var patbeforest1 = str1.substring (0, position); var patbeforest2 = str2.substring (0, position); if (checkmarchq (patbeforest1, patbeforest2)) {// alert (str1+'); 해당 하반기 var str1xbackposition = str1.length-str1xposition-1; var str2xbackposition = str2.length-str2xposition-1; var backposition = str1xbackposition> str2xbackposition? str2xbackosposition; if (backposition == 0) {alert (str1+'및'+str2+"mural"; str1.substring (str1.length-backosposition, str1.length); var patbackstr2 = str2.substring (str2.length-backposition, str2.length); if (checkmarchq (patbackstr1, patbackstr2)) {alert (str1+'및'+str2+"conf"); '+str2+"clict"); var str1xbackposition = str1.length-str1xposition-1; var str2xbackposition = str2.length-str2xposition-1; var backetosem = str1xbackosposition> str2xbackosposition? str2xbackosposition : str1xbackosposition; if (backposition == 0) {alert1 (str2+'); str1.substring (str1.length-backosposition, str1.length); var patbackstr2 = str2.substring (str2.length-backposition, str2.length); if (checkmarchq (patbackstr1, patbackstr2)) {alert (str1+'conflicts'+str2+"conflicts"); str2xposition! = -1) || (str1xPosition! = -1 && str2xPosition == -1)) // * {var strx = str1xPosition == -1? str2 : str1; // str1xposition == -1? str1 : // strx.leng.leng) <strx. strx.indexof ( '*'); if (position == 0) {// alert (str1+'및'+str2+"충돌 전반전"); var backposition = strx.length-position-1; if (backposition == 0) {alert (str1+''+str2+"); str1.substring (str1.length-backposition, str1.length); var patbackstr2 = str2.substring (str2.length-backposition, str2.length); if (checkmarchq (patbackstr1, patbackstr2)) {alert (str1+'conflicts'+str2+"conflicts"); str1.substring (0, 위치); var patbeforest2 = str2.substring (0, position); if (checkmarchq (patbeforeast1, patbeforeast2)) {// alert (str1+'및'+str2+"Conflict the Chorth Propection = strx.length-position-1; if (if) {alert (alert 1) 및 '+str2+"conflict");} else {var patbackstr1 = str1.substring (str1.length-backosposition, str1.length); var patbackstr2 = str2.substring (str2.length-backosposition, str2.length); if (checkmarchq (patbackstr1, patbackstr2)) {alert (alert (alert) 및 '+str2+"confrict");}}}}}}} else {if (checkmarchq (str1, str2)) {alert (str1+'및 '+str2+"conflict");}}}} 함수 checkmarchq (str1, str2) {var flagque = false; length; length are afe afe afe afe afe afe afe afe aft1.length; (var i = 0; i <str1.length; i ++) {if (str1.substr (i, 1)! = '? 충돌}}}} return flagque;}* 적어도 하나의 문자이며 위의 프로그램은 입력의 합법성을 확인하지 않습니다.
더 많은 JavaScript 구문을 보려면 "JavaScript 참조 자습서"및 "JavaScript Code Style Guide"를 따라갈 수 있습니다. 또한 모두가 wulin.com을 더 지원하기를 바랍니다.