Best Way To Show User Their Photo - php

Question

Okay so I was wondering what is the best way to show a user their own photo and if my way is safe or what should i change.
url:
http://localhost/project/everyone/myphoto.php?num=2
php code:
$user_id = $_SESSION['user_id'];
if (isset($_GET['num'])) {
$num = $_GET['num'];
if ($stmt = $dbconn->prepare("SELECT 1 FROM t_photos WHERE id ='$num' AND user_id ='$user_id' LIMIT 1")) {
$stmt->execute();
$stmt->store_result();
$rows = $stmt->num_rows;
if ($rows === 1) {
$stmt = $dbconn->prepare("SELECT url,uploaddate FROM t_photos WHERE id = ?");
$stmt->bind_param('i', $num); // Bind "$email" to parameter.
$stmt->execute(); // Execute the prepared query.
$stmt->store_result();
$stmt->bind_result($photopath, $uploadtime); // get variables from result.
$stmt->fetch();
} else {
$error2 = "Error 2";
require 'notfound.php';
die();
}
}
}
html & Php code :
<div id="pathwrap">
<div class="photowrap">
<?php if (isset($photopath)) {
echo '<img src="' . $photopath . '">';
} ?>
</div>
</div>

Answer 1

This is how I would do it with PDO and Exception style:
function requestCurrentUserPhoto(){
if( !isset($_GET['num']) ){
throw new Exception('Bad request. The generated link missing get prop num.');
}
if( !isset($_SESSION['user_id']) ){
throw new Exception('Bad request. The generated link linked to a guest.');
}
$sth = $dbh->prepare('SELECT url,uploaddate FROM t_photos WHERE id = :id AND user_id = :user_id LIMIT 1');
$sth->execute(array(
':id' => (int) $_GET['num'],
':user_id' => (int) $_SESSION['user_id']
));
$result = $sth->fetch(PDO::FETCH_ASSOC);
if( $result === false ){
throw new Exception('Bad request. The generated link linked to a non-existence photo or unauthorized user.');
}
//optional...
if( empty($result['url']) || empty($result['uploaddate']) ){
throw new Exception('Bad database table row. There is a invalid photo row in t_photos');
}
return $result;
}
This code should be safe. And it should also check if the code that is related got any errors.