有条件的ISSET问题

In a single product page, the product ID is assigned to the var $productID. The problem is when, staying in one of the fields of the form and push the ENTER key, the page crashes since the script can't find the ID again. So I would implement the script in the following way.

<?php
$product_id = $_GET['product_id'];
include '../sys/conn.php';
$risultato = mysqli_query ($conn, "

(mysql query)

") or die ("Query not valid: " . mysqli_error($conn));
mysqli_close($conn);
$row = mysqli_fetch_array($risultato);
?>

That I implemented in order that if $_GET is not set properly, it is set to 13321 (example value).

<?php

$product_id = $_GET['product_id'];

if(!isset($_GET['product_id']){ 
$product_id='13321';
}else{ 
include '../sys/conn.php';
$risultato = mysqli_query ($conn, "
(mysql query)

") or die ("Query not valid: " . mysqli_error($conn));
mysqli_close($conn);
$row = mysqli_fetch_array($risultato);
}
?>

The conditional clause looks like is not accepted and generates errors such like syntax or unexpected variables. Any help?

---VERSION 2----

<?php
$product_id = $_GET['product_id'];

if(!isset($_GET['product_id'])){ 
    $product_id='13321';}
    else 
{

include '../sys/conn.php';
$risultato = mysqli_query ($conn, "

(MYSQL Query)

") or die ("Query non valida: " . mysqli_error($conn));
mysqli_close($conn);
$row = mysqli_fetch_array($risultato)

}
?>

And the error is: Parse error: syntax error, unexpected '}' in /Applications/XAMPP/xamppfiles/htdocs/gest/pages/singolo_prodotto.php on line 66

You missed bracket:

if(!isset($_GET['product_id']){ 

Should be

if(!isset($_GET['product_id'])){