[PHP] include (byname.php?lang=fr);

I want to create a language option, so the resulting html is in French or English, depending. But passing variables like this doesn't work: ?php include (../_phpcode/byname.php?lang=fr); ? echo $lang; // 'fr' Warning: main(): Failed opening '../_phpcode/byname.php?lang=fr'

Re: [PHP] include (byname.php?lang=fr);

From: John Taylor-Johnston [EMAIL PROTECTED] I want to create a language option, so the resulting html is in French or English, depending. But passing variables like this doesn't work: ?php include (../_phpcode/byname.php?lang=fr); ? echo $lang; // 'fr' Warning: main(): Failed opening

Re: [PHP] include (byname.php?lang=fr);

Why didn't I think of that :) ? Thanks John, John John W. Holmes wrote: From: John Taylor-Johnston [EMAIL PROTECTED] I want to create a language option, so the resulting html is in French or English, depending. But passing variables like this doesn't work: ?php include