$.ajax work around

$.ajax work around

$.ajax({
                        type: "POST",
                        url: "index.php?mc=projects&sc=add",
                        data: {domain:domain, ownerid:ownerid, supervisorid:supervisorid, startDate:startDate, pname:domain},
                        dataType:"html",
                        cache: false,
                        success:
                       
                        alerts an error if domain has already a duplicate else returns true;
                      

            });

PHP:

$numRes = mysql_num_rows(mysql_query($query));
if($numRes > 0) {
      echo 'Failure';
}else {
      echo 'success';
}


Can you help me guys, i've been crazy doing this... I already tried using json but still failed..