在JSON响应中获取null

Question

我想加密数据并在解密后获取它。我正在使用

AES_ENCRYPT and  AES_DECRYPT functions of MYSQL

我试了很多指出错误，但我无法理解为什么我会得到无响应。请告诉我哪里出错了

这是我的代码

<?php
//Turn off all error reporting
//error_reporting(0);
ini_set ("display_errors", "1");

error_reporting(E_ALL);



 if(($_GET['action'])&&($_GET['username'])&&($_GET['key'])) {

$select = $_GET['action'];
$username =$_GET['username']; //no default
$key= $_GET['key'];
if($key=='dYrtr3GGPVuhSMvm'){
if($select=='select'){
        /* connect to the db */
$connect = mysqli_connect('localhost','root','')or die("Couldn't connect to database!");
mysqli_select_db($connect,'easy_sign') or die ("Couldn't find database");

 $key="x1y2z3";

$query ="SELECT *, AES_DECRYPT(path, '$key') AS decrypted_path, AES_DECRYPT(filename, '$key') AS decrypted_name FROM origin WHERE username ='$username' ";



$result = mysqli_query($connect,$query);
$numrows=mysqli_num_rows($result);
if($numrows!==0)
{
$post = array();
while($row = mysqli_fetch_array($result))
{
$post['path'] = $row['decrypted_path'];
 $post['filename'] = $row['decrypted_name'];
 $post['username'] = $row['username'];
 $post['date'] = $row['date'];
 }

 header('Content-type: application/json');
 echo json_encode($post); 


     }
    }
   }
  }

 ?>