try
eval("?>$footertext");
also take the code out of the function and try it, you may need to global some more
variables.
Paul Roberts
[EMAIL PROTECTED]
++++++++++++++++++++++++
----- Original Message -----
From: "Leif K-Brooks" <[EMAIL PROTECTED]>
To: <[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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