also take the code out of the function and try it, you may need to global some more 

Paul Roberts
----- Original Message ----- 
From: "Leif K-Brooks" <[EMAIL PROTECTED]>
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.

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