I understand your confusion.  The thing is that empty() and ! are two
completely different arguments.

if(empty($var)) is looking for: $var = '';

if(!$var) is looking for: $var = false; or $var = 0;

If $var is set to anything other than 0 or false then the ASCII value of
the string is (by definition) equivilant to an integer value of 1 or
more.  And thus is interpreted as true.

Hope that helps.

-Kevin

-----Original Message-----
From: David Johansen [mailto:[EMAIL PROTECTED]] 
Sent: Thursday, March 07, 2002 3:44 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Re: User accounts

Here's a little piece of code that gives me a weird problem:

<?php
if (!$logout)
{
   if ($_SESSION['loggedin'])
   {
      unset($_SESSION['loggedin']);
   }
}
?>


I have the session started and everything, but it gives me the following
error.
Warning: Undefined variable: logout in c:\inetpub\wwwroot\uslogin.php on
line 13
I've seen this used on several examples. I know that if I just use
empty($logout) then it'll work ok, but I just wanted to know why it's
used
in so many examples but doesn't work in my code. Is there some setting
that
I have set wrong or something? Thanks,
Dave

"David Johansen" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> I'm new to this php thing and I would like to set up a web page were
the
> users can login and edit their preferences and all that stuff. I have
the
> basic login stuff worked out and I was passing the username and
password
as
> a hidden input in the form, but then the password can be seen with
view
> source. I know that there's a better way to do this, so could someone
point
> me to a good tutorial or example on how I could make it so that the
user
> could login and logout and then I wouldn't need to be passing the
password
> all around like this. Thanks,
> Dave
>
>



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