php - How to count all rows from mysql but display only 2 results and count all results -

i trying display data database. more precisely images. problem when have extract more 2 images. want display first 2 images , print example added 49 images or how many images had extract mysql. script here:

public function getimages($id, $username){     global $con;     $stm = $con->prepare("select * image postid = :id , author = :username");     $stm->execute(array(":id" => $id, ":username" => $username));     $imagesnum = $stm->rowcount();     if($imagesnum == 1){         $image = $stm->fetch(pdo::fetch_obj);         echo '<div class="frame"><div class="frameoneimage"><img src="../upload/img/'.$username.'/'.$image->path.'"></div></div>';     } else if($imagesnum == 2){         echo '<div class="frame">';         $row = $stm->fetchall();         foreach($row $image){             echo'<div class="frametwoimages"><img src="../upload/img/'.$username.'/'.$image['path'].'"></div>';         }         echo '</div>';     } else {         // here want display 2 images write added (number of images extracted)     } } 

you can set variable increment , exit out of loop after 2 iterations, so:

else {     echo '<div class="frame">';     $row = $stm->fetchall();     $i = 0;     foreach($row $image){         echo'<div class="frametwoimages"><img src="../upload/img/'.$username.'/'.$image['path'].'"></div>';         if ($i >= 2) {             echo 'only first 2 results showing. ' . $imagenum . ' results have been omitted.';             break;         } else {             $i++;         }     } }