Re: Variable won't work in function, even when I global it?

try eval("?>$footertext"); also take the code out of the function and try it, you may need to global some more variables. Paul Roberts roberts_paul@bigfoot.com ++++++++++++++++++++++++ ----- Original Message ----- From: "Leif K-Brooks" <eurleif@buyer-brokerage.com> To: <php-general@lists.php.net> Sent: Monday, May 20, 2002 6:23 AM Subject: [PHP] Variable won't work in function, even when I global it? > On my website, I open my header (and footer) file with fileopen and > then eval() it. I know I should include them, but I knew nothing about > php when I did this. Anyway, I'm working on "groups" for my website. > They will be a kind of club. At the top of every group page, I include > a group function file. Among those functions is one that checks if the > user trying to go to the group is a member of it. If not, it gives them > an error message and exits. It is also supposed to eval() the footer > before exiting. My code is as follows: > > function checkmember(){ > global $groupid; > global $userinfo; > global $footertext; > if($groupid != $userinfo['groupid']){ > print "<b>Error:</b> you're not a member of this group."; > eval($footertext); > exit; > } > return; > } > > > The thing is, it doesn't seem to eval() the footer. I've even tried > getting $footertext out of the $GLOBALS array. I have also tried > printing $footertext to make sure it wasn't a problem with evel(). > Nothing has worked. When I did isset() on $GLOBALS['footertext'], it > worked however. Thanks to anybody who can help. > >