Ajax在发送之前就陷入困境

Question

我的Ajax登录表在发送前停留。

这是：

  

beforeSend：function（）

我试图显示错误，但没有任何错误，因为它一直停留在发送前。是因为我的PHP没有返回任何数据并影响JavaScript吗？

Ajax脚本：

$('document').ready(function()
{
/* validation */
$("#login-form").validate({
    rules:
    {

    },
    messages:
    {

    },
    submitHandler: submitForm
});
/* validation */
/* form submit */
function submitForm()
{
    var data = $("#login-form").serialize();

    $.ajax({
        type:'POST',
        url  : 'lgreg/post/login-post.php',
        data : data,
        beforeSend: function()
        {
            $("#error").fadeOut();
            $("#login-submit").html('Logining ...');
        },
        success :  function(data)
        {
            if(data===0){

                $("#error").fadeIn(1000, function(){


                    $("#error").html('Please Activate Your Acount!');

                    $("#login-submit").html('Login');

                });

            }
            else if(data=="success")
            {

                $("#login-submit").html('Redirecting...');
                var delay = 1000; 
                setTimeout(function(){ window.location = "http://moonxyweb.com/emailconfirmation"; }, delay);

            }
            else{

                $("#error").fadeIn(1000, function(){

                    //$("#error").html('Error occur, Please Contact Our Support!');
                    $("#error").html(data);
                    $("#login-submit").html('Login');

                });

            }
        }
    });
    return false;
}
/* form submit */

});

PHP脚本：

<?php
 ini_set('display_errors', 1);
 ini_set('display_startup_errors', 1);
 error_reporting(E_ALL);
 require_once 'dbconfig.php';

 if($_POST)
{
$user_name      = $_POST['user_name'];
$user_password  = $_POST['password'];
$lastlogin_date   = date('Y-m-d H:i:s');
$password   = password_hash( $user_password, PASSWORD_BCRYPT, array('cost' => 11));
$activation = "1";
try
{
    $stmt = $db_con->prepare("SELECT * FROM tbl_user WHERE user_name=:name AND activation=:activation");
    $stmt->execute(array(":name"=>$user_name , ":activation"=>$activation));
    $count = $stmt->rowCount();

    if($count==1){
        $stmt = $db_con->prepare("SELECT * FROM tbl_user WHERE user_name=:name AND user_password=:password");
        $stmt->execute(array(":name"=>$user_name , ":password"=>$user_password));
        $check_acc = $stmt->rowCount();
        if($check_acc == 1)
        {
            echo "success";
        }
        else
        {
            echo "wrong";
        }

    }
    else{
        echo "0";
    }

}
catch(PDOException $e){
    echo $e->getMessage();
}
}


?>