Displaying time logged in

[john9292]

Hello, im trying to display the amount of time the user has been logged in for on my site although its always one hour ahead e.g. if logged in for 20min its 1hour 20min. Tried to minus 360 from $start_time although this did not work.

Please help.
Thanks

<font size="1">
<?php echo "$status";?>
<br><?php echo "Logged in at:  ".$_SESSION['loginTime'];?></br>
<?php
$loginTime = $_SESSION['loginTime'];
$time = date("g:i a");

$start_time  =  strtotime($loginTime);
$end_time  =  strtotime($time);

$loggedInFor  =  ($end_time  - $start_time);
?>
<br><?php echo "Logged in for: " .date("g:i",$loggedInFor);?></br>
</font>

[cahva]

Dont use those date function to calculate. All you need is time() and a function to convert those seconds to human readable format. Ofcourse you can use that date function to show the logintime..

session_start();
if (!isset($_SESSION['logintime']))
        $_SESSION['logintime'] = time();

$timenow = time();
$loggedinfor = $timenow - $_SESSION['logintime'];


echo 'Logged in for: '.timeformat($loggedinfor);

// heres a function to format the seconds
function
timeformat($seconds) {
	$hours = floor($seconds/3600);
	$temp = $seconds%3600;
	$mins = floor($temp/60);
	$secs = floor($temp%60); 

return($hours.':'.$mins.':'.$secs);
}

[john9292]

Ok cheers for the help but now its something like 31000 hours ahead, lol.

I am using my own server with PHP installed - could it be the setup of that do you think?

[john9292]

I managed to resolve this by subtracting that figure I mentioned i.e.

$hours = floor($seconds/3600 - 322170);

For my website I also want to show which users are online and then update this when they log off or on.

I managed to display how many users are registered - I used a function to show how many rows were in the members table of the database.

Where would I start doing this or do you know any tutorials or guides on the internet?
Is this a complex task?

Thanks

[ClarkF1]

You could just add a logged on field to the database and update to yes or no when they log in or out.

You could then just query the database for those users where the loggedin field is set to yes

Or you could use a boolean field (ie. true or false)

[john9292]

I think ill try the first way you said. I am trying to pass email so it knows which login to change to no (this is the logout script). I managed to get the login script to change the login field in the database to yes.

Logout script not working - please give any hints why.

<?php
	session_start();
	require "connect.php";

$sql2 = mysql_query("UPDATE accounts SET login = 'no' WHERE email = '".$_SESSION['email']."'")
	or die ("Couldn't perform query $sql <br />".mysql_error());

header("Location: index.php");
exit();
?>

This is the page with the logout button on (trying to pass the email of the logged on person from here):

&nbsp;Registered members:</br>
<?php
	$sql = "SELECT email FROM accounts";
	$result = mysql_query($sql)
		or die ("Couldn't perform query $sql <br />".mysql_error());
	$nrows = mysql_num_rows($result);
	$email = mysql_fetch_row($result);
	echo $nrows;
	$_SESSION["email"] = $email;
?>

Thanks anyone

[WDPEjoe]

Care to give us the errors you're getting? It would sure help in trying to figure out what's going wrong...

[john9292]

Well I would give you some errors although it produces none i.e. when I comment out the redirect to index.php it just goes to the logout script with no errors.

As using a session does not work to pass the email, could anyone please suggest an alternative method or would the session idea work?

Cheers