Re: Why won't this work?

Eh.. probably should be: $result = mysql_query("SELECT id FROM ma_users WHERE user='$PHP_AUTH_USER'"); $row=mysql_fetch_row($result) print $row['id']; René René Moonen wrote:

Well if you *know* your query only returns one row or you are just interessed in the first row, just forget about the while construction $result = mysql_query("SELECT id FROM ma_users WHERE user='$PHP_AUTH_USER'"); $id=mysql_fetch_row($result) print $id; Good luck René John Wulff wrote:

Didn't work, but its a great place for me to start from. As far as the difference between a looping query and a non looping query, i understand the difference but not exactly how to go about writing a non looping query. This bit of code is looping, and i assume it dosen't need to be, how do i fix this? $result = mysql_query("SELECT id FROM ma_users where user='$PHP_AUTH_USER'"); while(list($id) = mysql_fetch_row($result)) { print($id); }

hmm... seems a MySQL topic. I suppose that your table is filled with correct data at some other point. So what you need is a count query that returns *one* result and not all records in that table. You do not want a *while* loop in your PHP script, because that would show a list of unique ips. So my gues would be: $result = mysql_query("SELECT COUNT(DISTINCT IP) AS ips FROM ma_counter"); $ips = mysql_fetch_row($result); pintf("Visitors = %d",$ips); Disclaimer: code not tested. You might want to check the MySQL manual for syntax details of the query. Good luck René John Wulff wrote:

You're a savior, don't know how i missed that. Thanks a bunch. Now, one more quick question if you don't mind. I've got this query at the bottom of my page for the purpose of a counter. But naturally i don't want a list of all the ips logged, i just want a count of how many unique ips there are in the table. How do i go about this?<?php $result = mysql_query("SELECT distinct ip from ma_counter"); while(list($ip) = mysql_fetch_row($result)) { print ("$ip"); } ?>

<snip>

      
            $mode = "entrance";
            if ($mode == "entrance") {

</snip> I suppose you added the first line ($mode = "entrance"; ) for testing purposes during debugging, but in order for the script to work you should now remove it, because now $mode will always have the value "entrance" good luck René