[RESOLVED] Anyone see a problem with this? Seem to be getting error constantly!

[bl4ckl4b3l]

<?
mysql_connect("xx","xx","xx") or die (mysql_error()); //connect to db
mysql_select_db("xx") or die (mysql_error()); //select db

if (($_REQUEST['id'])) {
	$id = $_REQUEST['id'];

$get = "SELECT * FROM sites WHERE id = $id";
$site = mysql_fetch_array($get);

$old_in = $site['in'];
$new_in = $site['in'] + 1;

$update = "UPDATE sites SET `out`='$new_in' WHERE id='$id'";
mysql_query($update) or die (mysql_error());

echo "redirecting to the website you requested..";
echo '<META HTTP-EQUIV=Refresh CONTENT="2; URL='.$url.'">';


}
else { 
	header("Location: http://www.google.com/"); 
}
?>

Keep getting the error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/filehunt/public_html/top/out.php on line 9. I'm positive the DB connecting info is right (xx'ed out)

[bl4ckl4b3l]

I can't believe I didn't see that error!

Forgot the MySQL_QUERY around the variable!!! Here's the correct..

<? 
mysql_connect("xx","xx","xx") or die (mysql_error()); //connect to db 
mysql_select_db("xx") or die (mysql_error()); //select db 

if (($_REQUEST['id'])) { 
    $id = $_REQUEST['id']; 

$get = mysql_query("SELECT * FROM sites WHERE id = $id"); 
$site = mysql_fetch_array($get); 

$old_in = $site['in']; 
$new_in = $site['in'] + 1; 

$update = "UPDATE sites SET `out`='$new_in' WHERE id='$id'"; 
mysql_query($update) or die (mysql_error()); 

echo "redirecting to the website you requested.."; 
echo '<META HTTP-EQUIV=Refresh CONTENT="2; URL='.$url.'">'; 


} 
else { 
    header("Location: http://www.google.com/"); 
} 
?>