无法加载服务器响应的资源，状态为404

Question

您好我在Chrome控制台中出现此错误：无法加载资源：服务器响应状态为404（未找到）http://localhost/wordpress/add_likes.php

js source

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.4/jquery.min.js"></script>

js代码是

function addLikes(id,action) {
    $('.demo-table #tutorial-'+id+' li').each(function(index) {
        $(this).addClass('selected');
        $('#tutorial-'+id+' #rating').val((index+1));
        if(index == $('.demo-table #tutorial-'+id+' li').index(obj)) {
            return false;   
        }
    });
     $.ajax({
    url: 'add_likes.php',
    data:'id='+id+'&action='+action,
    type: "POST",
    beforeSend: function(){
        $('#tutorial-'+id+' .btn-likes').html("<img src='http://localhost/wordpress/wp-content/uploads/2016/12/loaderIcon.gif' />");
    },
    success: function(data){
    var likes = parseInt($('#likes-'+id).val());
    switch(action) {
        case "like":
        $('#tutorial-'+id+' .btn-likes').html('<input type="button" title="Unlike" class="unlike" onClick="addLikes('+id+',\'unlike\')" />');
        likes = likes+1;
        break;
        case "unlike":
        $('#tutorial-'+id+' .btn-likes').html('<input type="button" title="Like" class="like"  onClick="addLikes('+id+',\'like\')" />')
        likes = likes-1;
        break;
    }
    $('#likes-'+id).val(likes);
    if(likes>0) {
        $('#tutorial-'+id+' .label-likes').html(likes+" Like(s)");
    } else {
        $('#tutorial-'+id+' .label-likes').html('');
    }
    }
    });
}

并且，是的，我正在使用wordpress。

我希望有人可以帮助我:(