JSON-encode several arrays

Question

I'm using JSON_encode to send data to a js file for tracks. A user could have one or several tracks so that's where the problem lies. Not sure how to organise my array to allow for several arrays. I can't use an array within an array since all the JSON data needs to be separated preferably so there will be something like;

track1 {ID:110232....}
track2 {ID:21402....}

What I have now works fine if there is just one track.

$ID = $_GET['ID'];
$result = mysql_query("SELECT * FROM tracks WHERE ID = '$ID' ORDER BY timestamp DESC");
while($row = mysql_fetch_array($result)){
    $T_ID = $row['T_ID'];
    $T_url = $row['url'];
    $T_name = $row['name'];
    $T_timestamp = $row['timestamp'];
    $arr = array('T_ID' => $ID,'T_name' => $T_name, 'T_url' => $T_url, 'T_timestamp' => $T_timestamp );

    echo json_encode($arr);
}

Please, don't use mysql_* functions in new code. They are no longer maintained and the deprecation process has begun on it. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Madara's Ghost
– Madara's Ghost, Commented Oct 29, 2012 at 13:06
Why does it need to be separated? It makes more sense to return one object with multiple nested objects. — Michael Berkowski
– Michael Berkowski, Commented Oct 29, 2012 at 13:06

Sirko · Accepted Answer · 2012-10-29 13:07:45Z

4

Why can't you use an (associative) array inside an container array like this:

$cont = array();
while($row = mysql_fetch_array($result)){
  $T_ID = $row['T_ID'];
  $T_url = $row['url'];
  $T_name = $row['name'];
  $T_timestamp = $row['timestamp'];
  $arr = array('T_ID' => $ID,'T_name' => $T_name, 'T_url' => $T_url, 'T_timestamp' => $T_timestamp );

  $cont[] = $arr;
}

echo json_encode($cont);

This results in a JSON structure like this, which keeps all your tracks in separate objects:

[ 
  {'T_ID': 1, 'T_name': 1, ... },
  {'T_ID': 2, 'T_name': 2, ... },
  {'T_ID': 3, 'T_name': 3, ... },
  ...
]

^{As noted in the comments you should switch to PDO or mysqli- functions, but this doesn't matter for the problem at hand.}.

answered Oct 29, 2012 at 13:07

Sirko

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2 Comments

Jacob Windsor Over a year ago

Thanks very much. I know this isn't part of the problem discussed but how would i return the results of all the JSON arrays in jQuery?

Sirko Over a year ago

@nbs189 jQuery has the parseJSON() function, which transforms the JSON string into a JavaScript object, which you can then use as you like.

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