What is the correct syntax to prevent error 1064?

I have mysql version 4.1 and have no problems with accessing the other databases I have setup, but there's one database, that I can't query, the line below is used:-

$db->open("SELECT * FROM requestlist
WHERE (ID = $requestID) LIMIT 1");

But when that line is reached, I get:-

status=err
SQL error=1064
SQL error string=You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') LIMIT 1' at line 2
SQL=SELECT * FROM requestlist WHERE (ID = ) LIMIT 1

what is the right syntax?

The full line of the code is:-

<?
####################
# Request Dedication
$requestID = $mainsong['requestID'];
$db->open("SELECT * FROM requestlist
WHERE (ID = $requestID) LIMIT 1");
if(($info = $db->row()) and (!empty($info['message'])))
require("dedication.php");
#===================
?>

Looks like $RequestID isn't being set correctly. Your error is coming from that missing value.

$requestID is set higher up in the script, the lines below are the beginning of the script.

$db->open("SELECT songlist.*, historylist.listeners as listeners, historylist.requestID as requestID, historylist.date_played as starttime FROM historylist,songlist WHERE (historylist.songID = songlist.ID) AND (songlist.songtype='S') ORDER BY historylist.date_played DESC LIMIT 6");
$history = $db->rows();
reset($history);

$db->open("SELECT songlist.*, queuelist.requestID as requestID FROM queuelist, songlist WHERE (queuelist.songID = songlist.ID) AND (songlist.songtype='S') AND (songlist.artist <> '') ORDER BY queuelist.sortID ASC LIMIT 2");
$queue = $db->rows();
reset($queue);

Should it be $queue['RequestID'] instead of $mainsong['RequestID'], then?