[SOLVED] Users Online (from scratch)

Ok i would like a script for users online. i am unfamilar with timestamps and such so any help would be greatly apprciated.

~ Chocopi

do a search on this -- for posts made by me -- there is one of these...  i made one that is on these forums.

is this the one you were talking about ?

<?php
if ($_SESSION['admin']['intime'] != '') {
if ($_SESSION['admin']['intime'] < time() - 1800) {
session_unset('admin');
$over = "1800";
echo "ERROR: you have been idle for more than 1800 seconds. Please login again";
}}

$_action = isset($_REQUEST['action']) ? $_REQUEST['action'] : '';

switch ($_action) {
default:
	if (!isset($_SESSION['admin'])) {
	if ($over != "1800") {
	echo 'Authorization Required.  Please log in.'
	}
	// login form html here
	} else {
		include of the page you wish to show
	}

break;

case login:
	if ($_REQUEST['act'] != "login") {
		if (isset($_SESSION['admin']) ) {
			// update session
		}
		if ($_SESSION['admin']['username'] != '') {
			// include of the page you want to view
		} else {
			if ($over != "1800") {
				echo "authorization required";
			}
			// login form include
		}
	} else {
		// the query to the database, (use LIKE BINARY for case sensitivity)
// also the cookie/session creation would go here
		if ($login == "true") {
		// show the actual page you want them to view 

		} else {
		// show the login form				
		}
	}
break;
}

?>

Thanks,

~ Chocopi

No that is a login script -- this is from a LONG LONG time ago -- let me see if i can find it on here for you.

-- edit:

after doing some searching I found my code...

<?php
$time = time() - 60 * 15;
$mems = "SELECT * FROM members WHERE last_activity <= $time";
$mems = mysql_num_rows(mysql_query($mems), 0);
$guest = "SELECT * FROM online_guests";
$guest = mysql_num_rows(mysql_query($guest), 0);
$tot = "SELECT * FROM members";
$tot = mysql_num_rows(mysql_query($tot), 0);
echo $mems." members online";
echo $guest." guests online";
echo $tot." total members";
?>

That code will generate a count of each who are online...  please note that I am no longer using this code, but it did work @ the time of conception -- you will need to modify it a little to fit your application.

<?php
$time = time() - 60 * 15;
$mems = "SELECT * FROM members WHERE last_activity <= $time";
$mems = mysql_query($mems);
$cnt = mysql_num_rows($mems);
for ($i = 0; $i < $cnt && $rows = mysql_fetch_assoc($mems); $i++) {
echo $rows['username'].', ';
}
?>

This here will give you a list of usernames who are currently online (within 15 minutes)...  This has not been tested, but by the looks of things should work without any problems....

<?php
$time = time() - 15 * 60
$mem = "SELECT inout FROM users WHERE username = '$user_name'";
$mem = mysql_query($mem);
$mem = mysql_result($mem, 0);
if ($mem == "in") {
echo '<img src="in.gif">';
} else {
echo '<img src="out.gif">';
}
?>

This here is to have a online/offline image next to a users name, in a forum setting this would need to be dramatically changed as this will only do a single user instead of all the users who have posted in a forum or whatever...

I hope that these 3 code samples help you in figuring out what you want and need to do, to get something working for your site.  And as I say to everyone, if you need any help I am only a PM away.

However, if you want somebody to completely make the code for you, please post your request in the freelancers area.