AJAX用户名可用性 [英] AJAX username Availability

问题描述

我已编写此代码来检查电子邮件的可用性.

I have writtin this code to check the email availability.

var email = $('#email_reg').val();
       if(email && email.length > 0)
       {
         if(!isValidEmailAddress(email))
         {
           isValid = false;
           $('#msg_email').html('Email is invalid').show();           
         }
         else
         {jQuery.ajax({
            type: 'POST',
            url: 'check_username.php',
            data: 'email='+ email ,
            cache: false,
            success: function(response){
                if(response == 1){
                    $('#msg_email').html('Email already Exists').show();
                    isValid=false;
                }
                else {
                    $('#msg_email').html('').hide();
                }
            }
        });

         }
       }
       else
       {
         isValid = false;
         $('#msg_email').html('Please enter email').show();
       }

PHP代码为

<?php
require_once('Connections/connection.php');
$username= mysql_real_escape_string($_REQUEST["email"]);

if (!$con)
{
    echo 0;
}
else {
    mysql_select_db($database_connection, $connection);
    $result = mysql_query("SELECT * FROM vendor_logiin WHERE username='" . $username . "'");
    $num = mysql_num_rows($result);
echo $num; //it will always return 1 or 0 since we do not allow multiple users with the same user name.
}
mysql_close();
?>

现在所有其他功能都可以正常工作，就像将其留空并提供错误的电子邮件格式一样.但是问题是当我提供已经存在的电子邮件ID时.它没有给出错误.我不知道出了什么问题.

Now all the others work well like when left it empty and give a wrong email format.But the problem is when i give an email Id that already exists. It didnot give error. I have no idea what is going wrong.

推荐答案

由于您未指定 dataType ，因此响应可能被视为文本或html，在这种情况下，这样做可能是明智的选择比较形式为字符串:

Since you didn't specify dataType the response is probably treated as text or html and in that case it might be wise to do the comparison as a string:

if (response == "1") {...}

而不是数字.或使用 parseInt(response，10)== 1 (如果将其与数字进行比较).

instead of a number. Or use parseInt(response, 10) == 1 if you compare it as a number.

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