Er... well I've seen a lot worse code than that but maybe you could use ...
if (isset($_SESSION['username']) !empty($_SESSION['username'])) {
echo 'Welcome '.$_SESSION['username'].', you are still logged in.';
}
else {
header... etc etc
}
Rich
-Original Message-
From:
$_SESSION['level'] to 2 but how can i check that with the above
line?
Thanks for your help,
Frank
- Original Message -
From: Rich Gray [EMAIL PROTECTED]
To: Frank Keessen [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, February 18, 2003 11:53 AM
Subject: RE: [PHP] $_SESSIONS and printing
]]
Sent: 18 February 2003 12:15
To: Rich Gray; [EMAIL PROTECTED]
Subject: Re: [PHP] $_SESSIONS and printing off..
Thanks,
But then another question;
if (isset($_SESSION['username']) !empty($_SESSION['username'])) {
echo 'Welcome '.$_SESSION['username'].', you are still logged in.';
I
: Rich Gray [EMAIL PROTECTED]
To: Frank Keessen [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Tuesday, February 18, 2003 1:25 PM
Subject: RE: [PHP] $_SESSIONS and printing off..
Um... how about...
echo 'Welcome '.$_SESSION['username'].', you are still logged in and your
authentication level
At 13:33 18.02.2003, Frank Keessen said:
[snip]
This is working fine. But i've set $_SESSION['level'] to value 2 (that's
done in the php file that calls this one). Now i want the above code to
check for the level and if is not equal to two than go to the
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