get online user list and remove unactive user from list

What language/database?

You would need to add a "last_activity" column to your database, and update it each time a user does an action (such as loads another page on your site), and then only select your count of "online users" who have an "last_activity" of sometime in the past 10 or 15 minutes.

you could also track your current users using getclicky.com just add the code they give you to your footer and your good to go. I think they track up to a 100,000 users at one time. Don't quote me on that though.

which language are you using?

--Create My Sql Table ---

CREATE TABLE IF NOT EXISTS `sessions` (
`id` int(11) NOT NULL auto_increment,
`id_session` varchar(256) NOT NULL,
`up_date` datetime NOT NULL,
`user` varchar(40) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;

---counter.php ---

$id_session=session_id();
$ro=mysql_query("SELECT * FROM sessions WHERE id_session ='".$id_session."'");
if($ro) {
if(mysql_num_rows($ro)>0) {
$upd=mysql_query("UPDATE sessions SET up_date=NOW(), user='".$_SESSION['username']."' WHERE id_session ='".$id_session."'");
}
else {
$into=mysql_query("INSERT INTO sessions (id_session, up_date, user) VALUES ('".$id_session."',NOW(),'".$_SESSION['username']."')");
}
}
$del=mysql_query("DELETE FROM sessions WHERE up_date<NOW()-INTERVAL '15' MINUTE");
function CountOnlineUsers($ifguest){
$q=mysql_query("SELECT COUNT(*)FROM sessions WHERE user".$ifguest);
if(mysql_num_rows($q)>0) {
return mysql_result($q,0);
}
}
$lang_stat=array(
'online_all'=>'Онлайн всего: ',
'online_guest'=>'Гостей: ',
'online_user'=>'Пользователей: '
);


---online.php

$guests=CountOnlineUsers("=''");
$users=CountOnlineUsers("!=''");
$all=$guests+$users;
echo $lang_stat['online_all'].$all."<br/>";
echo $lang_stat['online_guest'].$guests."<br/>";
echo $lang_stat['online_user'].$users."<br/>";