online/active users

pablodelapena

Im fairly new to this whole PHP shebang, and I may be slightly out of my depth, but I was wondering how to detect if a user is online or not.

I have a user list at the side of my site and currently it displays ALL registered members. Not so much a problem now but (hopefully) as the site gets bigger I want to only display online users.

I thought about using a MySQL thing, as in when they register their status column goes to online, but if they didnt actually log out they would appear as permenantly online! I dont really want that, or is it possible to have MySQL revert to the default after a set amount of time and have every page update the status column?

Any help would be great!

cheers
Pablo de la Pena

adamgeorge

Here's a common way of doing this:

first you need a database table to store all the current users online:

CREATE TABLE "session" (
"sUser" character varying(50) NOT NULL,
"iTime" integer DEFAULT '0' NOT NULL,
"sIP" character varying(15),
"bGuest" numeric(1,0),
CONSTRAINT pk_session_iID PRIMARY KEY (susername)
);

Then you need to include this function on every page. This will clear any old sessions, and update the data.

function setOnlineUsers(){
	$sIP = $_SERVER["REMOTE_ADDR"];

if(session_set()){
	$sMemberName = $_SESSION["sMemberName"];
	$bGuest = 0;
}
else{
	$sMemberName = $sIP;
	$bGuest = 1;
}

//Past half hour;
$iTimePast = time()-1800;
$result = mysql_query("DELETE FROM session WHERE iTime < " . $iTimePast);

if(!$result){
	trigger_error("Problem removing old Sessions from database", E_USER_ERROR);
	return false;	
}

$result = mysql_query("SELECT iTime FROM session WHERE sUser ='" . $sMemberName . "'");

if(!$result){
	trigger_error("Problem selecting Sessions from database", E_USER_ERROR);
	return false;	
}

$iTime = time();

if(mysql_numrows($result) == 0)
	$result = mysql_query("INSERT INTO session (sUser, iTime, sIP, bGuest) VALUES ('" . $sMemberName . "', '" . $iTime . "', '" . $sIP . "', '" . $bGuest . "')");
else
	$result = mysql_query("UPDATE session SET sUser = '" . $sMemberName . "', iTime= '" . $iTime . "', sIP = '" . $sIP . "', bGuest = '" . $bGuest . "' WHERE sUser = '" . $sMemberName . "'");

if(!$result){
	trigger_error("Problem creating Session", E_USER_ERROR);
	return false;	
}
else
	return true;
}

I think I misspelled some function names there too lol. You'll have to look them up in the documentation.

Anyway, all that function does is it first clears out all the old sessions (any older then 30 mins).

Then it checks wheter you're logged in or not. If so, it adds or updates the the session table so you're now included in it as a non-guest.

If you're not logged in, you're still added to the session table, but as a guest.

So then you can do queries on this table to see how many people have been on your site in the last 30 mins, including guests and non-guests.

Eg:


function whosOnLine(){
	mysql_query("select count(*) from session where bGuest = 0");
    list($iMembersOnLine) = mysql_fetchrow();

mysql_query("select count(*) from session where bGuest = 1");
list($iGuestsOnLine) = mysql_fetchrow();

echo "<p>There are currently " . $iGuestsOnLine . " guest(s) and " . $iMembersOnLine . " member(s) online</p>";

if(Session_set())
	echo "<p>You are logged in as <br><b>" . $_SESSION["sUserName"] . "</b></p>";
else
	echo "<p>You are not logged in. Please login or register to access all parts of the site</p>";
}

Hope that helps.

cya
-Adam 🙂

pablodelapena

wow, thanks! thats a bit much for me just now but ill trudge through it!

Great thats amazing!!!

cheers🙂