Posted by One on 10/03/06 01:59

Colin Fine wrote:
> One wrote:
> > I have a main.php file that calls a php navigation menu.
> >
> > I want to pass the menu file a parameter to tell it which menu to
> > display.
> >
> > Inside the main.php I have :
> > include "/home/ottadca1/public_html/includes/leftnav.php?menuid=dave";
> >
> > But the error is :
> > Warning:
> > main(/home/ottadca1/public_html/includes/leftnav.php?menuid=dave)
> > [function.main]: failed to open stream: No such file or directory in
> > /home/ottadca1/public_html/1menu.php on line 117
> >
> > Warning: main() [function.include]: Failed opening
> > '/home/ottadca1/public_html/includes/leftnav.php?menuid=dave' for
> > inclusion (include_path='.:/usr/lib/php:/usr/local/lib/php') in
> > /home/ottadca1/public_html/1menu.php on line 117
> >
> > However - if use the exact same line without passing a parameter it
> > works fine.
> > This works : include "/home/ottadca1/public_html/includes/leftnav.php";
> >
> > I don't understand. I've used this technique in the past.
> > Is there some sort of php configuration I am missing ???
> >
> > THANKS!
> >
>
> You seem to be confusing HTTP (or rather CGI) and PHP.
>
> The '?arg=value&arg=value' syntax is part of CGI, and is interpreted by
> CGI programs running typically on web browsers. While many CGI
> programs are written in PHP, the syntax has nothing to do with PHP: the
> program that interprets it might be in Perl, VB, Asp, or other languages.
>
> When you use
> include('filename');
>
> you are staying within PHP, telling it to read its some code from
> somewhere else.
>
> If you use
> include("http://url");
>
> you are telling your PHP programme to send an HTTP request to that
> server, and parse the result as PHP code. The server will send back just
> what you would get if specified that URL to your browser - which will
> not usually be PHP source.
>
> Colin

Big thanks to Colin and Jerry for such concise answers.

Thanks again - everything is working well and as expected.

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