Question

我想从数据库中删除一条记录，并且我想要使用ajax ..

所以我有一张桌子，我把它放在最后td这个：

<input type='image' src='./img/delete.png' onClick='deleteUser(".$utentiIscritti[$i][0].");' />

这是我的deleteUser功能：

function deleteUser(id){
                                    $.ajax({
                                        type:"post",
                                        url: "deleteUserAjax.php",
                                        data: {'id':id},
                                        success: function(data){
                                            console.log("OK");
                                            location.reload();
                                        },
                                       error: function(xhr, status, error){
                                           alert(xhr+"\n\n"+status+"\n\n"+error);
                                           console.log("KO");
                                       }
                                    });
                                }

这是我的php页面连接到db并删除记录：

<?php
$USERDB = "u";
$PASSWORDDB = "p";
$NAMEDB = "d";

$queryDeleteUser = 'delete from user where id = "'.$_POST['id'].'"';
$conn = mysql_connect("localhost", $USERDB, $PASSWORDDB)
        or die("Errore nella connessione al database: " . mysql_error());
mysql_select_db($NAMEDB) or die("Errore nella selezione del database: " . mysql_error());
mysql_query($queryDeleteUser) or die("Errore nella query: " . $queryDeleteUser . "\n" . mysql_error());
dbDisconnect($conn);

但我总是（从每个ajax请求）获得错误：

Failed to load resource: the server responded with a status of 500 (Internal Server Error)
iscritti.php:80

为什么???

Answer 1

您可以考虑两种解决方案。

您的代码有问题。尝试在它自己的上面执行它。只需在浏览器中调用它并检查结果！
您已为脚本指定了关系路径。 url: "deleteUserAjax.php"，尝试使用绝对路径并检查结果（url: "http://yourdomain.com/deleteUserAjax.php"）

Answer 2

也许让它更清洁：

HTML部分：

<input type='image' src='./img/delete.png' value='<?=$id?>'>

jQuery部分：

$(document).ready(function(){
    $("#delete").on("click", function(){
        var data = $(this).val();

        $.ajax({
            method: "POST",
            url: "page_you_handle_it.php?action=delete",
            data: {'id':id}
        }).done(function(data){
            //here you get response of your delete function!
        });
    });
});

PHP部分：

$host = "[HOST]"; //Like localhost
$user = "[USER]"; //Like root
$pass = "[PASS]"; //Like 123
$db = "[DB]"; //Like users

$con = mysqli_connect($host, $user, $pass, $db) or die ("Conntecting the Database gone wrong");

$id = $_POST['id'];

$query_str = "DELETE FROM user WHERE id = '$id'";
$query = mysqli_query($con, $query_str);

if (!$query) //Do not run the `$query` in the return parts because it already runs when you say `if (!$query)`
{
    echo 'Delete gone wrong';
}
else
{
    echo 'Delete succes!';
}

为什么我总是使用ajax错误进入我的php网页？

2 个答案: