2017-10-17 81 views
0

了解我,我可能做错了什么,但我看看如何解析数据从Javascript到PHP看起来像使用Ajax,它看起来不错,但它不想通过发送数据并更新它我的数据库。Ajax无法正确解析数据?

发生的只是一个对话框,里面没有任何东西。

AJAX:

$.ajax({ 
     url: 'updateCredits.php', 
     type: 'GET', 
     data: { 
      credits: totalcash 
     }, 
     success: function(data){ 
      alert(data); 
     } 
    }); 

PHP:

<?php 

require 'steamauth/steamauth.php'; 
include ('steamauth/userInfo.php'); 
include('mysql/config.php'); 

if(isset($_SESSION['steamid'], $_GET["credits"])) { 
    $credits = $_GET['credits']; 
    mysqli_query($db,"UPDATE set credits = credits + '".$credits."' WHERE steamid = '".$steamprofile['steamid']."'"); 
} else { 
echo 'An Error has occurred, this is either due to you not being logged in or something went wrong!'; 

} 

?> 
+1

你期望发生的?你没有在成功上回应任何事情。 另外你的代码是非常非常非常脆弱的[SQL注入](https://www.owasp.org/index.php/SQL_Injection) – ntzm

+0

侧面说明:您应该使用PUT如果更新操作 –

+0

这是工作正常。代码中的空警报意味着,在数据库中最可能没有错误和数据。 –

回答

0

你是不是从PHP返回任何东西片断成功

<?php require 'steamauth/steamauth.php'; 
include ('steamauth/userInfo.php'); 
include('mysql/config.php'); 
if(isset($_SESSION['steamid'], $_GET["credits"])) 
{ 
    $credits = $_GET['credits']; mysqli_query($db,"UPDATE set credits = credits + '".$credits."' WHERE steamid = '".$steamprofile['steamid']."'"); 
echo 'Success fully updated'; 
} else { echo 'An Error has occurred, this is either due to you not being logged in or something went wrong!'; } ?>