Question

我正在尝试使用ajax从javascript中将一些数据保存到数据库中。这是我的代码部分，我创建请求并将json格式的数据发送到save / save1.php：

urlPath = "save/save1.php";
var request = new XMLHttpRequest();
request.addEventListener('load',function() {
                console.log(this.responseText); 
 }); 
 idGame = "1";
 data = JSON.stringify({'idUser': idUser, 'idGame': idGame, 'score': score}); 

 request.addEventListener('error', function () {
                    document.write('There was an error with AJAX request!');
 });
request.addEventListener('load', this.onSuccess);

request.onreadystatechange = function(){
    if(request.readyState == 4){
            alert(request.responseText);
        }
  };

request.open("POST", urlPath);
request.setRequestHeader('Content-Type','application/json');
request.send(data);

这是我的save1.php文件：

<?php 
include '../database_connection.php';
var_dump($_POST);die(); //this was just for debuging and it prints that array is empty

if (isset($_POST['idUser']) && isset($_POST['idGame']) && isset($_POST['score']))
{ 
      $rez = $_POST['score'];
      $idGame = $_POST['idGame'];
      $idUser = $_POST['idUser'];
      $query = 'INSERT INTO scores (idUser, idGame, score) VALUES          ('.$idUser.','.$idGame.','.$rez.');';
      $sucess = mysql_query($query);
     if($sucess)
    {

         echo 'yes';

    }
   else        
   {
       sendJsonResponse(array(
             'error' => "not saved")); 
    }
}  
else
{
     sendJsonResponse(array(
         'error' => "not set"));  
 };
function sendJsonResponse($data) {
    header('Content-Type: application/json');
    die(json_encode($data));
}

我做错了什么？：（

编辑：我解决了我找到了解释：

我刚刚注意到你也在设置contentType ...为了工作，你将无法使用$ _POST var。相反，您需要从file_get_contents（“php：// input”）中获取“应该”为您提供原始JSON字符串的数据。从那里你应该能够使用PHP的json_decode将它变成一个真正的PHP对象。

所以我添加到我的save1.php文档中：

$data = json_decode(file_get_contents('php://input'), true);
if (isset($data['idUser']) && isset($data['idGame']) && isset($data['score']))
 { 
    $rez = $data['score'];
//and so on...

AJAX发布请求

0 个答案: