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Logout after time

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Join Date: Jul 2009

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Toxikr3

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Logout after time

Jul 18th, 2009

Hi guys, I have been working on a site and have managed to add a login script. I made an "status" script myself which tells people if a user is online or not.

I did that by using database(easiest way I could think of), when person logs in it changes the value in the table "user" in column "onoffstatus" to "Online" and when they click logout it changes it to "offline".

My problem is that if they close window or navigate away from it without clicking logout, it still says "Online". How can I fix it? I have been looking at timeouts and sessions a lot, and can't find what I am looking for.

Somethings to note:
I am using a free site host, and cannot access php.ini
I have tried to use .htaccess but don't know what to do...

I hope you guys can help me. Thanks.
-Toxikr3

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Join Date: Oct 2005

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humbug

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Re: Logout after time

Jul 19th, 2009

The easiest way to do this would be to set a column "lastactivity" to the current time every time the user loads any of your pages (just build it into your authentication script). You can then classify someone as "online" if they have been active in the past 5 mins.

Hope that's close enough to what you're after.

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Join Date: Jul 2009

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Toxikr3

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Re: Logout after time

Jul 19th, 2009

Thank you, I will try that.

EDIT: Well, I looked up time, and it is harder then I thought. Is it possible for you to show how the code can be used?
I have used:
$today = getdate();
print_r ($today):

It prints an array, which I do not know how to use. Also how would I check if they have been inactive for more then 5 mins and then change their logout status?

-Toxikr3

Last edited by Toxikr3; Jul 19th, 2009 at 4:32 pm.

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Join Date: Jul 2006

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pritaeas

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Re: Logout after time

Jul 19th, 2009

This may be an option:

http://www.phpfreaks.com/forums/index.php?topic=135013

"If it is NOT source, it is NOT software."
-- NASA

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Join Date: Jul 2009

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Toxikr3

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Re: Logout after time

Jul 19th, 2009

Thank you, I will try that now.
-Toxikr3

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Join Date: Jul 2009

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Toxikr3

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Re: Logout after time

Jul 19th, 2009

Gah!! It does not work. I have looked at the link and tried both of them, I am currently trying to use the simpler one:

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
$the_time = time() + (1*60);
if($_SESSION['timestamp'] < $the_time)
{
$_SESSION['timestamp'] = time();
echo "More time added "; //to see if it works
}else{
mysql_query("UPDATE user SET onoffstatus = 'Offline' WHERE username = '$userfinal'");
session_unset();   #Session_unset and Session_destroy
session_destroy();#Will remove all sessions.
header("location:index.php");#This code will send you back to the index page
}
echo "<font color=white>";
echo $the_time;
echo "   ";
echo $_SESSION['timestamp'];
echo "</font>";
$the_time = time() + (1*60);
if($_SESSION['timestamp'] < $the_time)
{
$_SESSION['timestamp'] = time();
echo "More time added "; //to see if it works
}else{
mysql_query("UPDATE user SET onoffstatus = 'Offline' WHERE username = '$userfinal'");
session_unset();   #Session_unset and Session_destroy
session_destroy();#Will remove all sessions.
header("location:index.php");#This code will send you back to the index page
}
echo "<font color=white>";
echo $the_time;
echo "   ";
echo $_SESSION['timestamp'];
echo "</font>";

The last part is to see if the values change(the do) and the "More time" is shown also, but when I wait about 2 min(with page open in background) and then go back to it and click a link, I am still logged in. I also tried closing window and going back after 2 mins and nothing...

Any help, or is there a way to do it with javascript...onunload execute logout? or window close...?

Thanks
-Toxikr3

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Join Date: Jul 2006

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pritaeas

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Re: Logout after time

Jul 19th, 2009

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
<?php
	if (! session_id()) {
		session_start();
	}
 
	if (! isset($_SESSION['timestamp'])) {
		$_SESSION['timestamp'] = time();
	}
 
	if (time() - $_SESSION['timestamp']  < 60)	{
		$_SESSION['timestamp'] = time();
		echo "More time added "; //to see if it works
	}
	else {
		mysql_query("UPDATE user SET onoffstatus = 'Offline' WHERE username = '$userfinal'");
		session_unset();               # Session_unset and Session_destroy
		session_destroy();             # Will remove all sessions.
		header("location: index.php"); # This code will send you back to the index page
	}
?>
<?php
	if (! session_id()) {
		session_start();
	}

	if (! isset($_SESSION['timestamp'])) {
		$_SESSION['timestamp'] = time();
	}
	
	if (time() - $_SESSION['timestamp']  < 60)	{
		$_SESSION['timestamp'] = time();
		echo "More time added "; //to see if it works
	}
	else {
		mysql_query("UPDATE user SET onoffstatus = 'Offline' WHERE username = '$userfinal'");
		session_unset();               # Session_unset and Session_destroy
		session_destroy();             # Will remove all sessions.
		header("location: index.php"); # This code will send you back to the index page
	}
?>

"If it is NOT source, it is NOT software."
-- NASA

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Join Date: Jul 2009

Posts: 27

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Toxikr3

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Re: Logout after time

Jul 19th, 2009

Thank you! Thank you! Thank you! It has worked, but I have one little concern =D

This only logs you out if you go to the page after the time has elapsed, so it only refreshes the values in the table after some "connection". I was wondering if it is possible to use AJAX to have a timer that keeps checking every minute to see if they are online. Is it possible to do that even if they are not on the page?

If you don't understand my problem:
I am trying to change the value in a table after a period of time EVEN IF the user isn't on the page... is that possible? I hope you understand.

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Join Date: Jul 2009

Posts: 27

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Toxikr3

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Re: Logout after time

Jul 19th, 2009

Just realized, I meant even if the user isn't on the website. Like if they have navigated away from the site, or closed their browser.

Sorry for double post.
-Toxikr3

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Join Date: Jun 2007

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kkeith29

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Nearly a Posting Virtuoso

Re: Logout after time

#10

Jul 19th, 2009

Use a cron job to set users offline after a certain amount of time.

Just have it run every 5-10 minutes and that will solve the problem without having to rely on javascript to do it (which is not a good thing to do).

Google is your friend.

Use [code] tags.

If you have found a solution to your problem, please mark the thread as SOLVED.