What about this:

$sql="select id,agent,host, DATE_FORMAT(time_in, '%M %d, %Y, %l:%i') AS
unixdate from logged_in WHERE userid='$current_user' ORDER BY id DESC
LIMIT 1";


I think this should work for your case.

Maxim Maletsky
www.PHPBeginner.com



-----Original Message-----
From: Jason Dulberg [mailto:[EMAIL PROTECTED]] 
Sent: marted́ 2 ottobre 2001 6.04
To: [EMAIL PROTECTED]
Subject: [PHP] mysql query for current id-1


This is kindof a weird question so bear with me as I try to explain.

I have a session table that gets updated when a user logs in/out. If
they don't logout, some info is left unchanged. I have a cron script
that takes care of the stray sessions so that's all good. In the
sessions table, there's a field "self_logout" which is Y when they
logout properly and N if the cron script removes their session.

What I'd like to do is when the user logs in next time, a search will be
made to look at that users last login session info. If they didn't log
out properly, a notice will appear.

So theoretically, I need to search for something like:

"users current id -1" or their last time of visit.

Here is the sql query that I have so far. But I think that I need to
remove the "self_logout='N'" because that doesn't show the actual last
result; rather it shows the last result where they didn't properly
logout.

$sql="select id,agent,host, DATE_FORMAT(time_in, '%M %d, %Y, %l:%i') AS
unixdate from logged_in WHERE (self_logout='N') AND
(userid='$current_user') ORDER BY id DESC LIMIT 1,1";

Here is my trimmed down table structure:

CREATE TABLE logged_in (
id tinyint(4) DEFAULT '0' NOT NULL auto_increment,
session varchar(100) DEFAULT '0' NOT NULL,
time_in timestamp(14),
time_out varchar(50) DEFAULT '-' NOT NULL,
self_logout char(1) DEFAULT 'N' NOT NULL,
KEY id (id)
);

Did that make any sense? To sum it all up, I just want to remind people
to click "logout" if they forgot the last time.

Thanks for any suggestions!!

__________________
Jason Dulberg
Extreme MTB
http://extreme.nas.net


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