vorrei richiamare delle foto in base al numero id_user che prendo dal db:

<?php
session_start();

?>
<html>
<head>


<body onLoad="setTimeout('document.step1.submit();', 3000);">
<form name="step1" action="step1.php" method="get">

<title>Step1foto</title>
</head>

<?php




$connessione=mysql_connect("localhost", "");
$selezione_db=mysql_select_db("guest_book", $connessione);

$query = mysql_query("SELECT id_user, data FROM messaggi ORDER BY data DESC limit 1");
while($riga = mysql_fetch_assoc($query))
{
$id_user = $riga['id_user'];
}
while($riga = mysql_fetch_assoc($query))
{
echo $riga["id_user"];
}


$nr_immagine = ($id_user - 833) * 10 + 1 + $offset;
$directory = "C:\Programmi\EasyPHP1-8\www\fotos";

$image_src = $directory . "image" . $nr_immagine . ".jpg";
echo "<img src=\"$image_src\">";

if($offset < 50) {
$offset++; }



?>
<?php

$_SESSION['user'] = $_GET['user'];
$_SESSION['gender'] = $_GET['gender'];
$_SESSION['year'] = $_GET['year'];
$_SESSION['month'] = $_GET['month'];
$_SESSION['day'] = $_GET['day'];
$_SESSION['country'] = $_GET['country'];

?>
</body>

</html>

pero' mi da diversi errori:


Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in c:\programmi\easyphp1-8\www\step1foto.php on line 24

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in c:\programmi\easyphp1-8\www\step1foto.php on line 28

Notice: Undefined variable: id_user in c:\programmi\easyphp1-8\www\step1foto.php on line 34

Notice: Undefined variable: offset in c:\programmi\easyphp1-8\www\step1foto.php on line 34

Notice: Undefined variable: offset in c:\programmi\easyphp1-8\www\step1foto.php on line 40

Notice: Undefined variable: offset in c:\programmi\easyphp1-8\www\step1foto