在使用PHP和Ajax进行服务器端验证时遇到问题 [英] Got problems in server side validation using PHP and ajax

问题描述

我期望有一个输出，但没有显示任何输出.我的代码中是否缺少内容或出了什么问题?在此先感谢

我的主页上有这行代码

I was expecting an ouput but it didn''t show any. Is there missing in my code or what''s wrong? Thanks in advance

I have this line of code in my main page

<form>
        <label for="text_email">E-mail:</label>
        <input id="text_email" type="email" >

        <script type="text/javascript" charset="utf-8" src="//ajax.googleapis.com/ajax/libs/jquery/1.8.0/jquery.min.js">
            $(document).ready(function(){
                $('#text_email').keyup(function(){
                    email($(this).val());
                });
            });
            function email(str){
                $.post('checkmail.php',{ email: str},
                function(data){
                    $('#check_email').html(data.returnValue);
                }, 'json');
            }
        </script>

        <label for="text_email" id="check_email"></label>
    </form>

这行代码

And this line of code

<?php
if(isset($_POST['email']))
{
    try
    {
        $pdo=new PDO('mysql:host=localhost;dbname=class;charset=utf-8', 'root');
    }
    catch(PDOException $e)
    {
        echo 'Failed: '.$e->getMessage();
    }

    $stmt=$pdo->prepare('SELECT email FROM class where email=:email LIMIT 1');
    $stmt->execute(array(':email'=>$_POST['email']));
    if($stmt->rowCount()>0)
    {
        $check='E-mail cannot be use.';
    }
    else
    {
        $check='E-mail can be use.';
    }

    echo json_encode(array('returnValue'=>$check));
}
?>

推荐答案

(document).ready(function(){

(document).ready(function(){

('#text_email').keyup(function(){ email(

('#text_email').keyup(function(){ email(

(this).val()); }); }); 功能email(str){

(this).val()); }); }); function email(str){

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