jQuery POST 上的 500（内部服务器错误）

Question

我正在向 PHP 脚本发送带有 jQ​​uery 库的 POST 请求，您将在下面找到脚本。结果我收到 500 内部服务器错误，有人知道如何解决吗？

我如何发送表单：

    $.post(
        "post.php",
        $( "#form" ).serialize()
    ).done(function( data ){
            generateFinishPage( data.link );
    });

我的 post.php

<?php
    error_reporting(E_ALL); 
    ini_set("display_errors", 1);
    $pdo = new PDO('...');
    
    $error = Array();

    if(empty($_POST['link']))
        $error [] = 'Er is geen link geselecteerd!';
        
    if(empty($_POST['sitename']))
        $error [] = 'Er is geen domein geselecteerd!';
        
    if(empty($_POST['image']))
        $error [] = 'Er is geen image geselecteerd!';
        
    $selectQuery = $pdo->query('SELECT * FROM trolls WHERE url_text="' . $_POST['url'] . '" AND url_domein="' . $_POST["sitename"] . '"');
    if($selectQuery->rowCount() == 1)
        $error [] = 'Dezelfde link bestaat al!';
        
    if(count($error) == 0){
        $query = $pdo->query('INSERT INTO trolls (created_time, swf_theme, url_text, url_domain, creator_ip, creator_name, creator_text, creator_email) 
            VALUES (:created_time, :swf_theme, :url_text, :url_domain, :creator_ip, :creator_name, :creator_text, :creator_email)');
        $query->execute(
            array(
                ':created_time'     =>  date('d-m-o'),
                ':swf_theme'            =>  $_POST['image'] . ".swf",
                ':url_text'             =>  explode('/', $_POST['url'])[1],
                'url_domain'            =>  explode('/', explode(".", $_POST['url'])[1])[0],
                ':creator_ip'           =>  $_SERVER['REMOTE_ADDR'],
                ':creator_name' =>  $_POST['creator_name'],
                ':creator_text'     =>  $_POST['creator_text'],
                ':creator_email'    =>  $_POST['creator_email']
            )
        );
        echo json_encode(array("link" => $_POST['url']));
    } else {
        echo json_encode(array("link" => array($error));
    }
    
    $pdo = null;

发送的表单数据：

网站名称：lnstagram.nl

链接：测试

图片：安妮

创建者名称：测试

creator_email:test@test.nl

creator_text:测试

谢谢！

Answer 1

检查您的选择查询....是“url_domein”还是“url_domain”？

你想在你的插入方法中做什么？

我看不到任何准备好的语句但是你调用了执行方法？

 $query->execute(
        array(
            ':created_time'     =>  date('d-m-o'),
            ':swf_theme'            =>  $_POST['image'] . ".swf",
            ':url_text'             =>  explode('/', $_POST['url'])[1],
            'url_domain'            =>  explode('/', explode(".", $_POST['url'])[1])[0],
            ':creator_ip'           =>  $_SERVER['REMOTE_ADDR'],
            ':creator_name' =>  $_POST['creator_name'],
            ':creator_text'     =>  $_POST['creator_text'],
            ':creator_email'    =>  $_POST['creator_email']
        )
    );

尝试像这样更改查询...使用准备

$query = $pdo->prepare('INSERT INTO trolls (created_time, swf_theme, url_text, url_domain, creator_ip, creator_name, creator_text, creator_email) 
        VALUES (:created_time, :swf_theme, :url_text, :url_domain, :creator_ip, :creator_name, :creator_text, :creator_email)');

使用 try catch 块进行查询以了解错误在哪里

try {
       // query here
       } catch (PDOException $pe) {
          die("Error:" . $pe->getMessage());
      }

Answer 2

直接访问你的 post.php，你会发现真正的错误。并且不要忘记在你的 post.php 顶部显示错误

error_reporting(E_ALL); nullini_set('display_errors', '1');