How to insert picture id to database - php

Question

I got a form with picture and submit button, which sends the user_id of the connected user and picture_id of the picture that they want to save into the table called pins in MySQL Database. I tried to pin different pictures from different users. Insert of user_id works but the picture_id stays the same(the last picture in pictures table) even though I click on submit below different pictures. Is it because of the form, or am I referring to it in a wrong way?
Im placing here the php code with the picture and the button to save it.
<?php
session_start();
require_once __DIR__.'/connect.php';
$sUsername = $_SESSION['txtUsername'];
try {
$stmt = $db->prepare('SELECT * FROM pictures INNER JOIN users ON users.user_id = pictures.user_id');
$stmt->execute();
$aRows = $stmt->fetchAll();
foreach ($aRows as $aRow) {echo
"<form action='display.php' method='POST'>
<div class=item><img src=images/$aRow->path>
<span class=caption>Picture id:$aRow->picture_id</span>
<input type='submit' name='submit'>
</div>
</form>";}
if (isset($_POST['submit'])) {
try {
$stmt = $db->prepare('SELECT * FROM users WHERE username = :sUserName');
$stmt->bindValue(':sUserName', $sUsername);
$stmt->execute();
$row = $stmt->fetch();
$iUserId = $row->user_id;
$stmt = $db->prepare('INSERT INTO pins VALUES( null, :iUserId, :iPictureId )');
$stmt->bindValue(':iUserId', $iUserId);
$stmt->bindValue(':iPictureId', $aRow->picture_id);
$stmt->execute();
}catch (PDOEXception $ex){
echo $ex;
}
}
}catch (PDOEXception $ex){
echo $ex;
}
?>