JSON/PHP not parsing correctly

Question

Firstly i'll point out the problem, basically inside my $.each(data, function (i, v) { code it's returning a bunch of variables which are echo'd from the PHP code which is using json_encode but when trying to place these variables or even alert() them it's showing undefined i've tried many ways to actually display the data, but it's always returning undefined and i don't know why since my code seems valid to my perspective.

The current javascript code i have is as follows

$.ajax({
        url: "functions/ajax.php",
        data: "func=auto",
        type: "GET",
        dataType: "json",
        success: function(data){
                $.each(data, function (i, v) {
                            var name = v['name'];
                            var player_id = v['id'];
                            alert(player_id);
                });
        },
        error: function (jqXHR, textStatus, errorThrown){
            console.log('Error ' + jqXHR);
        }
});

The current PHP code i have is as follows

$res = $DB->Query("SELECT * FROM `inventory` WHERE `account_id` = '$_SESSION[ID]'");
$data = array();
while($player = $DB->fetch_assoc()) {   
        $data['name'] = $player['name'];
        $data['id'] = $player['player_id'];
}
header('Content-type: application/json');
echo json_encode($data);

Just to sum the whole thing up, when using alert() on player_id it returns undefined in which obviously i want it to return the correct value

There is only one player in there, so don't .each() over it. Or run $data[] = array('name'=>$player['name'],'id'=>$player['player_id']); — Wrikken
– Wrikken, Commented Mar 13, 2013 at 21:47
Would that work because my while($player = $DB->fetch_assoc()) { is returning more than one result ? — Curtis Crewe
– Curtis Crewe, Commented Mar 13, 2013 at 21:50
Look at what $data is in your current script. It will only be last player in the list unless you append records to $data instead of overwriting it. — Wrikken
– Wrikken, Commented Mar 13, 2013 at 21:53

Marc B · Accepted Answer · 2013-03-13 21:52:58Z

6

You're not sending back a multi-level array, you're sending an array with just two elements. e.g. your JS code should be

    success: function(data){
               var name = data['name'];
               var player_id = data['id'];
             }

If that DB query IS supposed to send back multiple records, then you're building it wrong:

while($player = $DB->fetch_assoc()) {   
   $data[] = array('name' => $player['name'], 'id' => $player['player_id']);
}
echo json_encode($data);

and then your $.each() code should start working, as you ARE sending back a multidimensional array/object.

answered Mar 13, 2013 at 21:52

Marc B

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