本文通過源碼展示如何實現表單提交前，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">用戶名</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){
$(".yzmtips").html('Error:'+msg.toSource());
}
})
return false;
})
});
})(jQuery);
</script>
</body>
</html>
2、后端驗證碼檢測 checkcode.php
<?php
/**
* 用戶驗證碼驗證文件
* @Author:Zjmainstay
* @version : 1.0
* @creatdate: 2013-10-4
*/
session_start();
echo json_encode(array('status'=>(int)($_SESSION["CHECKCODE"] == $_POST['code'])));
exit;
演示地址：Ajax實現提交表單時驗證碼自動驗證演示
源碼下載地址：Ajax實現提交表單時驗證碼自動驗證
轉載請附帶本文原文地址：Ajax實現提交表單時驗證碼自動驗證，首發自 Zjmainstay學習筆記