本文通過源碼展示如何?表單提交前，驗證碼先檢測正確性，不正確則不提交表單，更新驗證碼。
1、前端代碼 index.html
<!DOCTYPE html><html><head> <title>驗證碼提交自驗證</title> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <meta http-equiv="Content-Language" content="zh-CN" /></head><body> <form action="doPost.php" method="POST"> <div class="row"> <label for="username">使用者名稱</label> <input type="text" name="username" id="username" /> </div> <div class="row"> <label for="mod-captcha-code">驗證碼</label> <input name="code" id="mod-captcha-code" size="6" class="zjcaptcha" style="width:80px" type="text"/> <img class="code-img" style="height:30px;width:80px;" src="createcode.php?t=0" onclick="this.src=this.src.substring(0,this.src.indexOf('?')+1)+Math.random();return false;" /> <script type="text/javascript" src="http://www.zjmainstay.cn/jquery/jquery-1.8.2.min.js"></script> <div class="yzmtips" style="color:red"></div> </div> <div class="row"> <input type="submit" value="提交" class="submitBtn"/> </div> </form><script>(function($){ $(document).ready(function(){ $(".submitBtn").click(function() { var obj = $(this); $.ajax({ url:'checkcode.php', type:'POST', data:{code:$.trim($("input[name=code]").val())}, dataType:'json', async:false, success:function(result) { if(result.status == 1) { obj.parents('form').submit(); //驗證碼正確提交表單 }else{ $(".code-img").click(); $(".yzmtips").html('驗證碼錯誤！'); setTimeout(function(){ $(".yzmtips").empty(); },3000); } }, error:function(msg){ $(".yzmtips").html('Error:'+msg.toSource()); } }) return false; }) });})(jQuery);</script></body></html>
2、後端驗證碼檢測 checkcode.php
<?php/*** 使用者驗證碼驗證檔案* @Author:Zjmainstay* @version : 1.0* @creatdate: 2013-10-4*/session_start();echo json_encode(array('status'=>(int)($_SESSION["CHECKCODE"] == $_POST['code'])));exit;
源碼下載地址：Ajax實現提交表單時驗證碼自動驗證
以上就是本文的全部內容，希望對大家的學習有所協助，也希望大家多多支援雲棲社區。
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