php - AJAX检查登录名是否存在

Question

我有一个登录框...

当用户开始输入时..我想检查输入的登录名是否存在于数据库中...

如果登录名存在，我将把登录按钮设置为事件...如果它不存在，我将把登录按钮设置为非事件...

当然，我需要AJAX通过PHP执行我的mySQL，但我不知道如何完成。 .

假设这是我的查询

<?php 
    $result = mysql_query("SELECT * FROM accounts WHERE name='mytextboxvalue'"); 
?>

怎么做

Answer 1

保持简单:

$(document).ready(function(){
    var Form = $('#myForm');
    var Input = $('input.username',Form)

    Input.change(function(event){
        Value = Input.val();
        if(Value.length > 5)
        {
            $.getJSON('/path/to/username_check.php',{username:Value},function(response){
                if(response.valid == true)
                {
                    Form.find('input[type*=submit]').attr('disabled','false');
                }else
                {
                     Form.find('input[type*=submit]').attr('disabled','true');
                }
            });
        }
    });
});

然后是 PHP 端..

<?php
//Load DB Connections etc.
if(!empty($_REQUEST['username']))
{
    $username = mysql_real_escape_string($_REQUEST['username']);

    if(isset($_SESSION['username_tmp'][$username]))
    {
        echo json_encode(array('valid' => (bool)$_SESSION['username_tmp'][$username]));
        die();
    }
    //Check the database here... $num_rows being a validation var from mysql_result
    $_SESSION['username_tmp'][$username] = ($num_rows == 0) ? true : false;

    echo json_encode(array('valid' => (bool)$_SESSION['username_tmp'][$username]));
    die();
}
?>