I’ve done a more simple way where instead of inserting the new random mug into the table, you could just have one entry and it updates every week.

I’ve got the code here but just check that the fields and table names are correct coz I was testing it out on one of my databases and may have made a mistake when changing it back to your table names…

$query = “SELECT id, pid,timestamp FROM randompic”;
$result = mysql_query($query) or die(‘Query failed: ‘ . mysql_error());
$line = mysql_fetch_array($result,MYSQL_ASSOC);

$id = $line['id'];
$randompic = $line['pid'];
$lateststamp = $line['timestamp'];
$daysleft = ceil((($lateststamp+604800)-time())/86400);

if((time()-604800)>$lateststamp){

$query = “SELECT pid FROM cpg132_pictures ORDER BY RAND() LIMIT 1″;
$result = mysql_query($query) or die(‘Query failed: ‘ . mysql_error());
$wee = mysql_fetch_array($result,MYSQL_ASSOC);

$newrandompic = $wee['pid'];
$newstamp=$lateststamp+604800;

$query = “UPDATE randompic SET pid=’$newrandompic’, timestamp=’$newstamp’ WHERE id=$id”;
$result = mysql_query($query) or die(‘Query failed: ‘ . mysql_error());
}

$query = “SELECT cpg132_pictures.id, cpg132_pictures.filename, randompic.pid FROM cpg132_pictures, randompic WHERE cpg132_pictures.id=randompic.pid LIMIT 1″;
$result = mysql_query($query) or die(‘Query failed: ‘ . mysql_error());
$poo = mysql_fetch_array($result,MYSQL_ASSOC);

…then echo all your stuff out.