What is the logic to update three images of same ID?

[tahakirmani]

I am trying to update 3 images of same product_id. I have a different product_table and Product_images table. The product_images table has 5 fields; image_id, product_id, name, images, ord. I am trying to update images for any specific product_id. The problem I was facing before was that it set the same 'Image' for all of the three 'images', that why I add 'ord' filed and it solver that problem. Now I have a new problem. I am unable to understand where to write update image' query. If I write it inside for loop then it runs 'Update Query' 6 times and if I write it outside for loop then its unable to find out $ord[] variable. Kindly tell the way to resolve this problem. Given below is the part of code I am working on.

/*
* ------------------- IMAGE-QUERY Test 002 --------------------
*/
if (isset ( $_FILES ['files'] ) || ($_FILES ["files"] ["type"] == "image/jpeg"))
{
$i = 1;
/*
* ----------------------- Taking Current Order_id ------------------------
*/

// $order_sql= "select MAX(ord) from product_images";
$order_sql = "SELECT ord from product_images where product_id=$id";
$order_sql_run = mysql_query ( $order_sql );
echo mysql_error ();
for($i = 1; $i <= $order_fetch = mysql_fetch_array ( $order_sql_run ); $i ++)
// while ($order_fetch= mysql_fetch_array($order_sql_run))
{)
echo 'ID ' . $order_id [$i] = $order_fetch [(ord)];
}

/*
* ----------------------- Taking Current Order_id ------------------------
*/
foreach ( $_FILES ['files'] ['tmp_name'] as $key => $tmp_name ) {
// echo $tmp_name."<br>";
// echo 'number<br>';
echo $image_name = $_FILES ["files"] ["name"] [$key];
$random_name = rand () . $_FILES ["files"] ["name"] [$key];
$folder = "upload/products/" . $random_name;
move_uploaded_file ( $_FILES ["files"] ["tmp_name"] [$key], "upload/products/" . $random_name );
echo '<br>';
echo $sql = "update product_images set name= '$random_name',images= '$folder' where product_id=$id andord=$order_id[$i]";
if ($query_run = mysql_query ( $sql )) {

echo '<br>';
echo 'Done';
}
else {
echo mysql_error ();
}
// $i=$i+1;
}
}


/*-------------------
IMAGE-QUERY Test 002
--------------------*/

[tahakirmani]

anyone for the help??