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Re: User login status? [message #174557 is a reply to message #174556] Fri, 17 June 2011 20:39 Go to previous messageGo to previous message
me is currently offline  me
Messages: 192
Registered: September 2010
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On 6/17/2011 4:06 PM, Luuk wrote:
> On 17-06-2011 21:13, Bill B wrote:
>> On 6/17/2011 2:53 PM, Luuk wrote:
>>> On 17-06-2011 13:25, Co wrote:
>>>> On 15 jun, 20:33, Luuk<L...@invalid.lan> wrote:
>>>> > On 15-06-2011 00:57, Jerry Stuckle wrote:
>>>> >
>>>> >
>>>> >
>>>> >
>>>> >
>>>> >
>>>> >
>>>> >
>>>> >
>>>> >> On 6/14/2011 4:22 PM, Co wrote:
>>>> >>> Hi All,
>>>> >
>>>> >>> I have this code that will change one field of the users record when
>>>> >>> he logs in:
>>>> >
>>>> >>> mysql_query("UPDATE myMembers SET last_log_date=now(), logged='1'
>>>> >>> WHERE id='$id' LIMIT 1");
>>>> >
>>>> >>> This way we can display all the users who are logged in on the main
>>>> >>> page.
>>>> >>> When the user logs of the record will be changed again:
>>>> >
>>>> >>> mysql_query("UPDATE myMembers SET logged='0' WHERE id='$id' LIMIT
>>>> >>> 1");
>>>> >
>>>> >>> However what if the user just forgets to log off, then he is gone but
>>>> >>> his status is still logged=1.
>>>> >>> Is there a way to make sure the user status is changed even when he
>>>> >>> forgets to log off?
>>>> >
>>>> >>> Marco
>>>> >
>>>> >> The web is stateless - there is no such concept as "logged in" or not.
>>>> >> And there is no way to know if the user is still reading your page,
>>>> >> has
>>>> >> gone to coffee (or lunch), closed the browser or even shut off the
>>>> >> computer. So there is no way to determine how many people are
>>>> >> logged in
>>>> >> or not - and sites which claim to do that are pretty much guessing.
>>>> >
>>>> >> The best you can do is create your own custom session handler which
>>>> >> stores the data in a database instead of the file system. Then you
>>>> >> can
>>>> >> assume as long as the session is active, the user is logged in. When
>>>> >> the session expires, the user is logged out. Still not real accurate,
>>>> >> but the best you can do.
>>>> >
>>>> >> Just a warning - custom session handlers can be complex and pretty
>>>> >> much
>>>> >> beyond what you can do in a newsgroup.
>>>> >
>>>> >> Otherwise, Luuk's response is as good as any. And it will be about as
>>>> >> accurate as anything. You need a cron job to handle the deletion
>>>> >> of rows.
>>>> >
>>>> > You could also check if users are still logged on, just before
>>>> > gathering
>>>> > the list of logged on users...
>>>> > (and when someone logs in)
>>>> >
>>>> > This way your database will not give the correct 'state' to you, but to
>>>> > the web-users it will, because you put all the changes in there before
>>>> > you show anything to the web-users...
>>>> >
>>>> > --
>>>> > Luuk
>>>>
>>>> Luuk,
>>>>
>>>> I like the idea but I don't know exactly how to check if users are
>>>> still logged in?
>>>>
>>>> Marco
>>>
>>> i meant to logg the user off when they have not clicked for a defined
>>> number of minutes (after 10, or 15 minutes, or so....)
>>
>>
>> Any thoughts on how this would work with an<object> that points to a
>> remote swf? The person might be running the swf for some time without
>> causing any activity on the page that contains the<object>.
>>
>> Bill B
>>
>>
>
> swf has nothing to do with PHP.... ;)
>
> I'm not into swf-stuff.....

The page in question looks like this:
<html>
</html?
<body>
<?php
get $gamepath $game from $_SESSION variables
?>
<object width type
data=http:domain/$gamepath/$game/game.swf
param name = some stuff
</object>
</body>
</html>

So, the question, as I understand it, is how to monitor this page for
the inactivity you referred to earlier, when the user could be playing
the swf based game and not move off the page for a while.

If, for example:

<?php
while $seconds_since_last click <= $inactivity_limit;
do nothing
endwhile;
logoff_user();
?>

would not the focus on the <object> cause the inactivity limit to be
reached [potentially] before the person is finished with the game?

Unless I'm way off, seems like a PHP question to me.

Bill B
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