即使请求成功，Ajax也会返回404

Question

问题是我在ajax请求的数据库表中使用了几行代码。当我尝试确保更新成功时，它返回404标题，我转到网络工具和我看到404如下面的屏幕截图所示：

这是我对js和php

的代码

PHP：

<?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");
}

?>

JS：

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);
});

这很奇怪，我真的不知道是什么引起404标题，我们你们可以提供任何帮助，我会非常感激。