Question

I managed to output the column ‘resort’ in the Json array, but I need ‘country’ too, as well as ‘aantal’. Have no idea how to do that. Can someone please help me?

if ($numrows < 1 && strlen($sq) > 3)
{
        $sql = "SELECT resort, country, COUNT(image) AS aantal FROM sx_cam
          LEFT JOIN sv_orte ON sv_cam.res_id = sv_orte.res_id
          WHERE sound=soundex('$sq') and (status < 1) GROUP BY resort order by aantal desc";
        $result2 = mysql_query($sql) or die(mysql_error());
        $numrows = mysql_num_rows($result2);
        $suggest = 2;
}

$items = array();

while($row = mysql_fetch_assoc($result2)){
 $items[$row['resort']] = $row['resort'];
}

foreach ($items as $key=>$value) {
 echo strtolower($key)."|$value\n";
}

Answer 1

You’re building the array the wrong way. Once you get the array right, it is as simple as making a call to json_encode

I’m not entirely sure how you want your json to look, but something like this should get you started

$items = array();

while($row = mysql_fetch_assoc($result2)){

    //first we build an 'object' of the current result
    $item['country'] = $row['country'];
    $item['resort'] = $row['resort'];

    //now push it on the array of results
    $items[] = $item;
}

echo json_encode($items);

Once you get the above code working, you can tweak the PHP array to change the structure of the JSON to suit your needs.