如何使用json_encode作为基于我的代码的json返回

Question

Here, I want to encode php return in json output.I'm so confused to implement on there. So, how the correct way I have to do.

index.html

    $(function(){
         $.ajax({
            type: 'GET',
            url: "profile.php", 
            success: function(resp){
                var username = JSON.parse(resp).username;
                var profile = JSON.parse(resp).profile;
                $('.test').html(username+profile );

            }
         });
       });

profile.php

<?php
  require_once('class.php');
?>

<?php    
if ($user->is_logged == 1) {
    $txtuser = '';
    if (empty($D->me->firstname)) $txtuser = $D->me->username;
    else $txtuser = $D->me->firstname;

    if (empty($D->me->avatar)) $txtavatar = 'default.jpg';
    else $txtavatar = $D->me->avatar;
}
?>

<?php
 echo json_encode(array('username' => '{$C->SITE_URL.$D->me->username}', 'profile' => '{$txtuser}' ));
?>

Answer 1

clean it out this way, and let me know if you still have issues:

<?php
  require_once('class.php');

  if ($user->is_logged == 1) {
    $txtuser = '';
    if (empty($D->me->firstname)) $txtuser = $D->me->username;
    else $txtuser = $D->me->firstname;

    if (empty($D->me->avatar)) $txtavatar = 'default.jpg';
    else $txtavatar = $D->me->avatar;
  }

  $arr = array(
     "username" => $C->SITE_URL . $D->me->username,
     "profile" => $txtuser
  );

  echo json_encode($arr);
?>

in your Ajax response use console.log(resp) to see any errors in the console.