i need simple json in php from mysql like below given format

Question

i am use this code what change in this code

<?php
   $mysqli = new mysqli ( 'mysql2.000webhost.com', 'a8124593_local', 'root123', 'a8124593_test');
 // Check if album id is posted as GET parameter
   $myq = $mysqli ->query ( 'SELECT id, url, name FROM babyapp' );
  while ( $myr = $myq->fetch_assoc () ) {
   $array = (array("items"=>(array(
  'id' => $myr['id'],
  'name' => $myr['url'],
  'songs_count' => $myr['name'],
   ))));
  $arr_result[] = $array;
 }
 echo json_encode($arr_result);

?> result: http://pixelerbricks.net78.net/youtubeapp/babyapp.php

but my format wrong pls help me i would like to result in this format like. what change in my code for result give in below like format

 { 
 "items": 
[
     {
     "id":1,
             "name":"China",
     "url":"http://www.androidbegin.com/tutorial/flag/china.png"
     }, 

     {
     "id":2,
             "name":"india",
     "url":"http://www.androidbegin.com/tutorial/flag/china.png"
     }
    ]

}

possible duplicate of [Pretty-Printing JSON with PHP](http://stackoverflow.com/questions/6054033/pretty-printing-json-with-php) — Quentin, Jan 20 '14 at 08:16
Your PHP code up and the result you desire are not at all related, please be clear. — AyB, Jan 20 '14 at 08:17
from where do you get the rank, country, population and flag data ? — Ionut Flavius Pogacian, Jan 20 '14 at 08:20
output: http://pixelerbricks.net78.net/youtubeapp/babyapp.php but i need result display like above which have already given in qustion — Abhishek Maheshwari, Jan 20 '14 at 08:24
why do you change the entire json structure after I gaved you the solution? -1 for that — Ionut Flavius Pogacian, Jan 20 '14 at 08:31
sorry but json structure same only name change thank you very much my problem solved — Abhishek Maheshwari, Jan 20 '14 at 08:34

score 1 · Accepted Answer · answered Jan 20 '14 at 08:27

<?php
   $mysqli = new mysqli ( 'mysql2.000webhost.com', 'a8124593_local', 'root123', 'a8124593_test');
 // Check if album id is posted as GET parameter
   $myq = $mysqli ->query ( 'SELECT id, url, name FROM babyapp' );
  while ( $myr = $myq->fetch_assoc () ) {
   $array["items"][] = (array(
  'id' => $myr['id'],
  'name' => $myr['url'],
  'songs_count' => $myr['name'],
   ));
 }
 echo json_encode($array);

i need simple json in php from mysql like below given format

1 Answers1