使用 jquery 从 php 检索的中断数组

我从名为 check_num.php 的 php 文件中检索数组：-

check_num.php

 <?php
include 'config.php';
session_start();
$VALUE = $_SESSION["some_session_variable"];
if(isset($_POST['default'])){
$ert = "SELECT * FROM table_name WHERE something = '$VALUE' ORDER BY p_id ASC ";
$qty = mysql_query($ert);
$fgh = mysql_num_rows($qty);
$ertz = "SELECT something, COUNT(something) FROM table_name WHERE something = '$VALUE' 
     AND something >= 1 GROUP BY p_id ORDER BY p_id ASC";
$qtyz = mysql_query($ertz);
$tyui = mysql_num_rows($qtyz);
$data = array(
        "post"   => $fgh,
        "likes"  => $tyui
    );
    echo json_encode($data);
  } else {
 echo "0";
  }
?>

现在来了jquery部分：-

  <script>
   $(document).ready(function(){
    setInterval(function(){
    var def = "one";
    $.post("check_num.php", {'default': def },  function(response){
     if(response != 0){
        document.getElementById("total_array_count").innerHTML = response;
        //document.getElementById("total_like_count").innerHTML = response.likes;
        //document.getElementById("total_post_count").innerHTML = response.post;
      ------------------OR THIS Method-----------------
                     var my_array = response;
                     //var post_number = my_array["post"];            
        document.getElementById("total_array_count").innerHTML = my_array;
        //document.getElementById("total_post_count").innerHTML = '<b>'+post_number+'</b>';
        }
        else {

        document.getElementById("total_array_count").innerHTML='Error occured !';
        }
            });
           },2500);
           });
          </script>

现在收到的输出是{"post":10,"likes":1}，它是一个数组。但是当我response.post或my_array["post"]访问数组值时，返回的值undefined.

我经历了这个：- http://www.w3schools.com/js/tryit.asp?filename=tryjs_array_object

还有这种方式：- jQuery .val（） 返回未定义的单选按钮

跟着它，但没有成功！

请纠正我的错误.