RE: [PHP] Function / String Problem

2001-01-19 Thread Chris Mulcahy
PHP Mailing List Subject: [PHP] Function / String Problem Okay, what is wrong with this: echo "...are you logged in: $LoggedIn()"; // outputs:...are you logged in: () Function is: function LoggedIn(){ global $HTTP_POST_VARS if ($HTTP_POST_VARS["Login"] ==

Re: [PHP] Function / String Problem

2001-01-19 Thread Marcelo Gulin
Hi Karl! try... echo "...are you logged in: " . $LoggedIn() . ""; regards Marcelo Gulin "Karl J. Stubsjoen" escribió: > > Okay, what is wrong with this: > >echo "...are you logged in: $LoggedIn()"; > // outputs:...are you logged in: () > > Function is: > function LoggedIn(

Re: [PHP] Function / String Problem

2001-01-19 Thread Shawn Blaylock
Try this: echo "...are you logged in: " . LoggedIn() . ""; The reason it wasn't working is because it thought LoggedIn was a variable and () was part of the echo statement. "Karl J. Stubsjoen" wrote: > > Okay, what is wrong with this: > >echo "...are you logged in: $LoggedIn()"; > //

RE: [PHP] Function / String Problem

2001-01-19 Thread Boget, Chris
>echo "...are you logged in: $LoggedIn()";> // outputs:...are you logged in: () > > Function is: > function LoggedIn(){ > global $HTTP_POST_VARS > if ($HTTP_POST_VARS["Login"] == "PHPIsCool") > {return "YES";} > else > {return "NO";} > } Change your

[PHP] Function / String Problem

2001-01-19 Thread Karl J. Stubsjoen
Okay, what is wrong with this: echo "...are you logged in: $LoggedIn()"; // outputs:...are you logged in: () Function is: function LoggedIn(){ global $HTTP_POST_VARS if ($HTTP_POST_VARS["Login"] == "PHPIsCool") {return "YES";} else {return "NO";} }