Ok, im trying to make a "User Online:" script, but one that prints out the amount of users on a page... e.g.

Users online : 17
/index.php 6
/chat.php 3
/forums.php 8

I am using another script, but I like to rewrite my own. THis is what I have done so far.

<?php
$link_id = mysql_connect ("localhost", "######", "######");
mysql_select_db ("online", $link_id);

$query = mysql_query ("SELECT DISTINCT ip FROM useronline", $link_id);

while ($results = mysql_fetch_array ($query)) {
  $query = mysql_query ("SELECT file FROM useronline WHERE ip = '" . $results["ip"] . "'", $link_id);
}
?>

I know the first query is set, becuase I can loop through $result["ip"] and get it to print out the 10 IPs held in the database.

But, on running this comand... I get the following error message.

Notice: Undefined index: ip in C:\Apache\htdocs\online.php on line 8

Line 8:   $query = mysql_query ("SELECT file FROM useronline WHERE ip = '" . $results["ip"] . "'", $link_id);

Ive tried doing.....

$temp_ip = $results["ip"];
$query = mysql_query ("SELECT file FROM useronline WHERE ip = '" . $temp_ip . "'", $link_id);

With no avail.

What I did find interesting is that. when using this script...

<?php
$link_id = mysql_connect ("localhost", "######", "######");
mysql_select_db ("online", $link_id);

$query = mysql_query ("SELECT DISTINCT ip FROM useronline", $link_id);

while ($results = mysql_fetch_array ($query)) {
  echo ($results["ip"] . "<br>");
  //$query = mysql_query ("SELECT file FROM useronline WHERE ip = '" . $results["ip"] . "'", $link_id);
}
?>

It prints out all the 10 IP's in the DB, but notice the commented out "//$query"... only when its commented out will it echo $results["ip"]!

This is a defualt install of the latest PHP binary for windows (APache 2 - in ISAPI mode).

Hope this helps solve mine and anyone elses problems.