Create array in php and use Jquery get

Question

I'm trying to make a call to the database and recieve a "toplist" that is limited to 10 in PHP. Here I need to make an array and give it back to the Jquery with $.get().

But it's failing in recieving all the data with this code, How can I make this to be recieved and then do a "for each" in the Jquery for all the data that is being sent back from the PHP?

Jquery:

$.get("core.inc.php?get_toplist=1",
    function(data) {
        var json = JSON.parse(data); 
        alert(json['name']);
});

PHP:

if ($_GET['get_toplist']) {
    $get_top_list = mysql_query("
            SELECT * FROM ".$DBprefix."Submissions 
            WHERE status='1' 
            ORDER BY points DESC LIMIT 10");

    $i = 0;
    $arr = array();
    while ($top_list = mysql_fetch_array($get_top_list)) {   
        $arr[] = array(
            'position' => $i++,
            'name' => $top_list['name'], 
            'points' => $top_list['points']
        ); 
    }

    echo json_encode($arr);

}

@MichaelBerkowski It's calling json_encode(), so it's returning JSON. The header doesn't have any effect on the data that's returned. — Barmar
– Barmar, Commented Oct 22, 2013 at 13:30
In your browser's console, you should at least have seen some output returned from the AJAX request, though it is likely not in the format jQuery is expecting. — Michael Berkowski
– Michael Berkowski, Commented Oct 22, 2013 at 13:30
@MichaelBerkowski No, it doesn't. jQuery just returns the raw data as a string, and JSON.parse() parses it. — Barmar
– Barmar, Commented Oct 22, 2013 at 13:31
You get undefined because the data returned is a multi dimensional array and name is inside an inner node — Mina
– Mina, Commented Oct 22, 2013 at 13:32

Barmar · Accepted Answer · 2013-10-22 13:30:47Z

2

You're returning an array, but your Javascript code is treating it as a single element. Try:

alert(json[0].name);

If you want a foreach, you can do:

$.each(json, function(i, el) {
    console.log(el.name + " score is " + el.points);
});

answered Oct 22, 2013 at 13:30

Barmar

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Comments

developerCK · Accepted Answer · 2013-10-22 13:33:45Z

1

Friend your json will be like .

[{"position":1,"name":"name1","points":"points"},{"position":2,"name":"name2","points":"points"},   {"position":3,"name":"name3","points":"points"},....]

so you can not directly access it like ur way.

make a loop then access it.

$.each(data,function(i,val){
  alert(val.name);
  });

this will work.

answered Oct 22, 2013 at 13:33

developerCK

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Comments

Harish Singh · Accepted Answer · 2013-10-22 13:36:09Z

your php code is not proper

if ($_GET['get_toplist']) {

          $get_top_list = mysql_query("
          SELECT * FROM ".$DBprefix."Submissions 
          WHERE status='1' 
          ORDER BY points DESC LIMIT 10");

          $i = 0;  
          $arr = array();                                                                                                    
          while ($top_list = mysql_fetch_array($get_top_list)) {   

          $arr[] = array(
                    'position' => $i++,
                    'name' => $top_list['name'], 
                    'points' => $top_list['points']
                    ); 
          }
        }

echo json_encode($arr);

}

just add '}' before echo json_encode($arr);

also you are getting an array change in alert(json[0]['name']);

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