Online/Offline indicator

[tommmmm]

I'm looking to build an online/offline indicator for a site I'm working on. The login script uses sessions, nothing fancy just creates a default session with session_start(); there is no logout script/page so users can't officially logout in a very structured way.

Is there a simple way I can keep this system and have an online/offline indicator for each user? It would be nice if it was fairly accurate if it were 10/15 minutes out that wouldn't be a problem.

Cheers

[jignesh1]

i think wat u could do is that if the session is set then u could have a user online u could have that on a header of the page

example

if(isset($_SESSION['username'])){
echo "user is online";
}else{
echo "user is offline";
}

that is the simple way
i know of it

[tommmmm]

Sorry should have explained more, the indicator would go on the persons profile page...like myspace style. So the online/offline indicator would have to be stored in a mysql users table.

Cheers

[Piranha]

I would add a field in the database named "last_activity" and is a datetime field. Whenever the person that have the account do something it is set to the current time. When you want to check if the person is online you can check if the person have the last_activity less than 15 minutes ago. If it is less than 15 minutes the person is online, otherwise not. Note that 15 minutes is only a suggestion, you can use any time interval you want.

[rajug]

I think the session is not the solution. But as Piranha said a field in the database should have added and change the flag or just store there timestamp of the user login and a flag into another field will be better solution although i have not used this yet but it will be useful for me too if someone says a perfect way or algorithm for this.

[tommmmm]

I've started using the function

function date_diff($str_start, $str_end){ 
$str_start = strtotime($str_start); // The start date becomes a timestamp 
$str_end = strtotime($str_end); // The end date becomes a timestamp 
$nseconds = $str_end - $str_start; // Number of seconds between the two dates 
$ndays = round($nseconds / 86400); // One day has 86400 seconds 
$nseconds = $nseconds % 86400; // The remainder from the operation 
$nhours = round($nseconds / 3600); // One hour has 3600 seconds 
$nseconds = $nseconds % 3600; 
$nminutes = round($nseconds / 60); // One minute has 60 seconds, duh! 
$nseconds = $nseconds % 60; 
return $ndays.$nhours.$nminutes.$nseconds;
}

and then

$now = date("Y-m-d G:i:s"); 
if(date_diff($online, $now)<15){
$online_status='online';
}else{
$online_status='offline';
}

where $online is the database datetime field from when the user logged in.