Javascript 即使请求成功，Ajax也会返回404

问题是，我正在使用ajax请求对数据库表中的几行进行更新，当我尝试确保更新成功时，它返回404头，我转到网络工具，看到404，如下面的屏幕截图所示：

这是我为js和php编写的代码

PHP:

---

同意@charlietfl，else语句在AJAX响应中毫无意义。@charlietfl不，这不是问题所在，我刚刚试过，我这样做是为了重定向用户，如果他们热链接这个文件404将意味着找不到
ajax/update\u ads.php
——开发者工具网络选项卡中POST请求的整个URL是什么？那么请求发送到的实际URL是否与您期望的路径相同？可能需要更改为
var url='/ajax/update_ads.php'如果ajax目录位于根目录下level@charlietfl是的，更新查询进行得很顺利，我查看了数据库，我给值添加了种子，但是响应是404！这没有任何意义
<?php 
include("../../db_connect.php");

if(isset($_POST['ads_left'])) {

    $ads_left = mysqli_real_escape_string($connect , $_POST['ads_left']);
    $ads_right = mysqli_real_escape_string($connect , $_POST['ads_right']);
    $ads_top = mysqli_real_escape_string($connect , $_POST['ads_top']);
    $ads_bottom = mysqli_real_escape_string($connect , $_POST['ads_bottom']);

    $ads_left = htmlentities($ads_left);
    $ads_right = htmlentities($ads_right);
    $ads_top = htmlentities($ads_top);
    $ads_bottom = htmlentities($ads_bottom); 

    $ads_array = array();
    $ads_array[0] = $ads_top;
    $ads_array[1] = $ads_right;
    $ads_array[2] = $ads_bottom;
    $ads_array[3] = $ads_left;

    for ((int) $i = 0 ; $i<4 ; $i++) {
        $j = $i+1;
        $ins_ads_query = "UPDATE ads SET code = '$ads_array[$i]' WHERE id = '{$j}'";
        $result_ads = mysqli_query($connect , $ins_ads_query);
    }

} else {
    header("Location: ../index.php");
}

?>

var adsTop = document.getElementById('top');
var adsRight = document.getElementById('right');
var adsBottom = document.getElementById('bottom');
var adsLeft = document.getElementById('left');

var adsSubmit = document.getElementById("ads_submit");

function updateAds(e) {

    e.preventDefault();

    adsSubmit.innerHTML = "Saving <i class=\"fa fa-circle-o-notch fa-spin\" style=\"font-size:24px\"></i>";

    var http = new XMLHttpRequest();
    var url = 'ajax/update_ads.php';

    var params = 'ads_top='+adsTop.value+'&ads_right='+adsRight.value+'&ads_bottom='+adsBottom.value+'&ads_left='+adsLeft.value;
    http.open('POST', url, true);

    //Send the proper header information along with the request
    http.setRequestHeader('Content-type', 'application/x-www-form-urlencoded');

    http.send(params);

    http.onreadystatechange = function() {//Call a function when the state changes.
        if(http.readyState == 4 && http.status == 200) {
            adsSubmit.innerHTML = "Saved <span class=\"glyphicon glyphicon-floppy-saved\"></span>";
        }
    }
}


adsSubmit.addEventListener('click' , function(e) {
    updateAds(e);
});