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

I know that. Footertext prints out the footer for my site. The bottom of my layout. Also, as I said, I even tried printing $footertext. Thanks for trying to help, though. Miguel Cruz wrote:

I don't know what $footertext looks like, but I can tell you that just eval()ing a chunk of text won't have any apparent effect. When you eval() code in a string, that code has to either return() something, or have direct effects on its own (such as calling functions or setting values). eval() != echo. miguel On Mon, 20 May 2002, Leif K-Brooks wrote:

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.