It's after a variable called $LoggedIn.
You'll have to do:
echo "<br><br>...are you logged in: ".LoggedIn()."<br><br>";
hth
Chris Mulcahy
[EMAIL PROTECTED]
-----Original Message-----
From: Karl J. Stubsjoen [mailto:[EMAIL PROTECTED]]
Sent: Friday, January 19, 2001 11:17 AM
To: PHP Mailing List
Subject: [PHP] Function / String Problem
Okay, what is wrong with this:
echo "<br><br>...are you logged in: $LoggedIn()<br><br>";
// outputs: ...are you logged in: ()
Function is:
function LoggedIn(){
global $HTTP_POST_VARS
if ($HTTP_POST_VARS["Login"] == "PHPIsCool")
{return "YES";}
else
{return "NO";}
}
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