if else stament in php

Good Day. Here my question: i want to display the result if the count get from the SQL is more than 0, if the count is 0, then the result will not display. Here the current output after adding if else statement in my php :

    {"result":[]}

This is the output before adding if else statement in my php:

 { "result": [
    {"day":"1","count":"0"},
    {"day":"2","count":"2"},
    {"day":"3","count":"0"},
    {"day":"4","count":"0"}....

I want the result only like tis :

    {"result":[
      {"day":"2","count":"2"}
    ]}

Here my php code:

define('HOST','xx');
define('USER','xx');
define('PASS','xx');
define('DB','x');


//$month = $_POST['month'];
//$year = $_POST['year'];

$month = 10;
$year=2016;

$con = mysqli_connect(HOST,USER,PASS,DB);

$result = array();
 for ($day=1 ; $day<=31 ; $day++) {
     $sql = "select count(entryID) from Entry where
         EXTRACT(DAY FROM(entryTime)) = '$day'  AND
         EXTRACT(MONTH FROM(entryTime)) = '$month' AND 
         EXTRACT(YEAR FROM(entryTime))= '$year'";
     $res = mysqli_query($con,$sql);

if($row[0] >0)
     $day2 = (string)$day;
     while($row = mysqli_fetch_array($res)) {
         array_push($result,
         array('day' =>$day2,
            'count' => $row[0]
         ));

         }
}

 }


 echo $appear . json_encode(array("result"=>$result));



mysqli_close($con);

?>