Warning: mysql_fetch_array() expects parameter 1 to be resource, string given...
im tryin to put class inside code if $dayNumber and $dan are equal
i tried few ways but all are complicated since i have to call some functions this end up as "simple" solution but i dont know where did i get wrong PS that should be possible?
else {
echo "<td width='40'";
$moj_id = $_SESSION['id'];
$query_event = mysql_query("SELECT * FROM events WHERE id_user='$moj_id' GROUP BY 'event_date'") ;
$test = mysql_fetch_array('$query_event');
$date_event = $test['event_date'];
sscanf($date_event, "%d-%d-%d", $godina, $mjesec, $dan);
if ($dan==$dayNumber && $mjesec==$month && godina==$year)
{
echo " class='radni'";
}
echo"><a href='day_info.php?dan=",$dayNumber,"&mj=",$month,"&gd=",$year,"'>" . $dayNumber . " </a></td>\n";
$dayNumber++;
}