PHP - how do I return JSON in a specific format I need?

Question

I have a PHP file which currently returns JSON in these two ways:

If an error happens, I do this:

$post_data = array('error' => "no_member_id");
echo json_encode($post_data);

and if there is no error, and I need to return data in JSON format, I do this:

if (mysql_num_rows($result) > 0 )
{
         $rows = array();
         while($r = mysql_fetch_assoc($result))
         {
             $rows[] = $r;
         }

         echo json_encode($rows);
}

But what I really need to do is return the data in a format like this:

{"result":"ok", data :[{"data1":"value1", "data2":"value2"}]}

or this:

{"result":"error", data :[{"error":"no_id"}]}

Could someone please help me understand how to do that?

Thanks!!

Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. — Kermit
– Kermit, Commented Mar 13, 2013 at 13:20
@AarolamaBluenk thank you. How long do you think I have before these statements stop working in my code? :) — Genadinik
– Genadinik, Commented Mar 13, 2013 at 13:24
I don't think it matters; you should switch as soon as possible. — Kermit
– Kermit, Commented Mar 13, 2013 at 13:31
@AarolamaBluenk of course timing matters. I can't just rewrite all my code ASAP. I am curious how much time I realistically have. — Genadinik
– Genadinik, Commented Mar 13, 2013 at 13:51
When PHP 5.5 is released and used, you will get an E_DEPRECATED notice. — Kermit
– Kermit, Commented Mar 13, 2013 at 14:05

Emissary · Accepted Answer · 2013-03-13 13:43:11Z

2

echo json_encode( array( "result" => "ok", "data" => $rows ) );

instead of

echo json_encode($rows);

edited Mar 13, 2013 at 13:43

Emissary

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answered Mar 13, 2013 at 13:22

Pedro Cordeiro

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Comments

djjjuk · Accepted Answer · 2013-03-13 13:25:00Z

1

first, stop using mysql built in functions. they will be deprecated.

try this:

$result = 0;
$json = array(
 'result' => 'ok',
 'data'   => array()  
);
if (mysql_num_rows($result) > 0 )
  while($r = mysql_fetch_assoc($result)) {
    $json['data'][] = $r;
  }
} else {
  $json['result'] = 'error';
  $json['data'] = array('error' => "no_member_id");
}
echo json_encode($json);

answered Mar 13, 2013 at 13:25

djjjuk

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Comments

Vijay Verma · Accepted Answer · 2013-03-13 13:25:05Z

1

Just add result key with specific value in both array:

echo json_encode(array("result" => "ok", "data" : $rows));

and

echo json_encode(array("result" => "error", "data" : $post_data));

answered Mar 13, 2013 at 13:25

Vijay Verma

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