JSON output using PHP

Question

HI i need to output a JSON object for consuming in iphone

i am able to output like

{"feed":{"id":"1","player":"player1"}}
{"feed":{"id":"1","player":"player2"}}
{"feed":{"id":"2","player":"player3"}}

The code :

$query = "SELECT id,player FROM MyVideos";
$result = mysql_query($query,$link) or die('Errant query: '.$query);

$players[] = array();
if(mysql_num_rows($result)){
while($player = mysql_fetch_assoc($result)){
$players[] = array('player'=>$player);
echo json_encode(array("feed"=>$player));            
}
}

But i need to output some thing like this

{"feed":
{"id":"1","player":"player1"},
{"id":"1","player":"player2"},
{"id":"2","player":"player3"}
}

Can anyone please help me with this.

Thanks,

user1114 · Accepted Answer · 2011-08-30 00:42:14Z

7

The output you posted isn't valid JSON, you need to put brackets around the items in feed:

{"feed": [
    {"id":"1","player":"player1"},
    {"id":"1","player":"player2"},
    {"id":"2","player":"player3"}
]}

You should loop through your results and build an array of your feed items, and then output it all at once, like this:

$feed_items = array();

if (mysql_num_rows($result)) {
    while ($player = mysql_fetch_assoc($result)){
        $feed_items[] = $player;
    }
}

echo json_encode(array("feed" => $feed_items));

edited Aug 30, 2011 at 0:42

answered Aug 30, 2011 at 0:18

user1114

1

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2 Comments

UnlimitedMeals Over a year ago

Thanks Jeremy I tried this now the output is like {"feed":[[],{"id":"1","player":"Sourav Ganguly"},{"id":"1","player":"Sourav Ganguly"},{"id":"2","player":"Rahul Dravid"}]} not sure y the extra [[] are coming after feed.

user1114 Over a year ago

@UnlimitedMeals: Oops! The first like should have been $feed_items = array(); instead. That should fix it! If it does, and this solves your problem, consider clicking they grey checkmark on the left side of this answer to mark it as the "accepted answer".

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