我一直坚持这个问题好几天了。我用阿贾克斯组Web开发技术来调用从服务器的PHP文件。看来，成功的方法没有被调用。这是我的code：

I have been stuck with this problem for days already. I used Ajax group of web development techniques to call the php file from the server. It appears that the success method was not called. Here is my code:

function handleLogin() {
var form = $("#loginForm");    
//disable the button so we can't resubmit while we wait
//$("#submitButton",form).attr("disabled","disabled");
var e = $("#email", form).val();
var p = $("#password", form).val();

console.log("click");
if(e != "" && p != "") {
    //var str = form.serialize();
    //alert(str);
    $.ajax({ 
                     type: 'POST', 
                     url: 'http://prefoparty.com/login.php', 
                     crossDomain: true,
                     data:  {email: e, password :p},
                     dataType: 'json', 
                     async: false,

                     success: function (response){ 
                        alert ("response"); 
                        if (response.success) { 
                            alert("you're logged in");
                            window.localStorage["email"] = e;
                            window.localStorage["password"] = md5(p); 
                            //window.localStorage["UID"] = data.uid;           
                            window.location.replace(main.html);
                        } 
                        else {

                            alert("Your login failed");
                            //window.location("main.html");
                        }


                     },
                     error: function(error){
                         //alert(response.success);
                        alert('Could not connect to the database' + error);
                        window.location = "main.html";
                    }
    }); 
}
else {
    //if the email and password is empty
    alert("You must enter email and password");

}
return false;
}

在PHP中，我用了一个典型的MySQL的电话，当我运行谷歌Chrome浏览器的这个文件。它正确地返回的JSON。这里是我的PHP：

In php, I used a typical MySQL call and as I run this file from Google Chrome browser. It returned the JSON correctly. Here is my php:

            <?php
            require_once('includes/configinc.php');

            $link = mysql_connect(DB_HOSTNAME, DB_USERNAME,DB_PASSWORD) or die("Could not connect to host.");
            mysql_select_db(DB_DATABASE, $link) or die("Could not find database.");

            $uname = $_POST['email'];
            $password = $_POST['password'];



            $sql = "SELECT * FROM User_Profile WHERE Email = '$uname' AND Password = 'md5($password)'";
            $result=mysql_query($sql);
            $num_row = mysql_num_rows($sql);
            $row=mysql_fetch_array($result);


            if (is_object($result) && $result->num_rows == 1) {
            $response['success'] = true;

            }
            else
            {
            $response['success'] = false;
            }

            echo json_encode($response);

             //echo 'OK';

            ?>

请检查我的code和指出哪里我做错了。

Please check my code and point out where I did wrong.

感谢大家提前：）

推荐答案

添加

 header("access-control-allow-origin: *") 

要顶你的PHP页面将解决您的访问跨域请求的问题

to the Top of your PHP page will solve your problem of accessing cross domain request