problem with query -php/json

Question

i have this code, and the output is: {"round (avg (salary), 0)":"1750"}

function tableOne() {

    $query = mysql_query("select round (avg (salary), 0) from worker, formation_area where id_formation_area=1") or die(mysql_error());
    $row = mysql_fetch_assoc($query);
    $result = $row;

    echo json_encode($result);
}

tableOne(); 

?>

but i need something like this: {1750}.

Any help?

try doing a print_r($result); so you can see the data you're working with. — jrn.ak
– jrn.ak, Commented May 18, 2011 at 18:28

Naftali · Accepted Answer · 2011-05-18 18:28:00Z

4

Try this:

function tableOne() {

    $query = mysql_query("select round(avg (salary), 0) as `round_avg` from worker, formation_area where id_formation_area=1") or die(mysql_error());
    $row = mysql_fetch_assoc($query);
    $result = $row['round_avg'];

    echo json_encode($result);
}

tableOne();

Remember json needs a something: something_else so you might get an error here since now there is no array being encoded

answered May 18, 2011 at 18:28

Naftali

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