Question

我想在服务器上测试文件时收到错误（500内部服务器错误）。一切都与mamp（本地）一起运行良好，我没有收到任何错误。这是带错误的代码。

<?php
    include_once('../classes/places.class.php');
try
{
    $oPlace = new Places();
    $oPlace->Street = $_POST['place'];
    $oPlace->HouseNumber = $_POST['number'];
    $oPlace->Name = $_POST['Name'];
    if($oPlace->placeAvailable())
    {
        $feedback['status'] = "success";
        $feedback['available'] = "yes";
        $feedback["message"] = "Go ahead, street is available";
    }   
    else
    {
        $feedback['status'] = "success";
        $feedback['available'] = "no";
        $feedback["message"] ="De zaak " . "'" . $_POST['name'] . "'". " is reeds op dit adres gevestigd." ;;
    }
}
catch(exception $e)
{
    $feedback['status'] = "error";
    $feedback["message"] =$e->getMessage();

}
header('Content-type: application/json');
echo json_encode($feedback);
?>

Answer 1

它是什么版本的PHP？

如果在5.2之前，则需要安装JSON PECL软件包。

如果是5.20或更高版本，则必须检查PHP是否在没有--disable-json选项的情况下编译。

Answer 2

$feedback["message"] ="De zaak " . "'" . $_POST['name'] . "'". " is reeds op dit adres gevestigd." ;;

应该更像

$feedback["message"] ="De zaak " . "'" . $_POST['name'] . "'". " is reeds op dit adres gevestigd." ;

添加太多的半冒号有时会抛出错误

Answer 3

<?php
include_once('../classes/places.class.php');
/* This if for debugging */
foreach ($_GET as $k => $v) $_POST[$k] = $v;
// Access in your browser: pathToFilePHPCalled.php?place=SomePlace&number=14&Name=MyName
$feedback['data'] = $_POST;
/* This if for debugging */

$feedback = array();
try
{
    $oPlace = new Places();
    $oPlace->Street = $_POST['place'];
    $oPlace->HouseNumber = $_POST['number'];
    $oPlace->Name = $_POST['Name']; // Make sure this is $_POST['Name'] and not $_POST['name'] this might be your error
    if($oPlace->placeAvailable())
    {
        $feedback['status'] = "success";
        $feedback['available'] = "yes";
        $feedback["message"] = "Go ahead, street is available";
    }   
    else
    {
        $feedback['status'] = "success";
        $feedback['available'] = "no";
        $feedback["message"] ="De zaak " . "'" . $_POST['name'] . "'". " is reeds op dit adres gevestigd." ;
    }
}
catch(Exception $e)
{
    $feedback['status'] = "error";
    $feedback["message"] =$e->getMessage();

}
header('Content-type: application/json');
echo json_encode($feedback);
?>

500内部服务器错误php - ajax

3 个答案: