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

I won't send the whole value of $footertext, it's too long. Yes, it contains the contents of the footer file. I have tried displaying it before the function and it works fine. Martin Towell wrote:

what's that value of $footertext? is it the actual contents of the footer file? have you tried displaying the contents of $footertext just before you call checkmember()? does it contain what you're expecting? -----Original Message----- From: Leif K-Brooks [mailto:eurleif@buyer-brokerage.com] Sent: Monday, May 20, 2002 3:23 PM To: php-general@lists.php.net 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.