为什么我总是在php网页中使用ajax获取错误信息？

我想从数据库中删除一条记录，要做到这一点，我想使用ajax

因此，我有一个表，在其中我将最后一个

td

放入：

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

这是我的php页面，用于连接到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);

---

<>为什么？？？< P/>

你可以考虑两种解决方案。< /P>

你的代码有缺陷。试着自己执行它。只需在浏览器中调用它并检查结果

您已经为脚本指定了一个关系路径<代码>url:“deleteUserAjax.php”，尝试使用绝对路径并检查结果（

url:http://yourdomain.com/deleteUserAjax.php“

）

---

也许可以让它更干净：

HTML部分：

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

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!';
}

---

你试过在没有ajax的情况下运行脚本吗？@AdRock我不明白你想要什么告诉我。。。如果我删除ajax，我的脚本中就没有任何内容了…@BrianCoolidge你能告诉我在哪里吗？快来吧。我会用PDO tho

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

$(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!
        });
    });
});

$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!';
}