Try putting GLOBAL infront of the var to give it the greater scope
----- Original Message -----
From: "Chris Anderson" <[EMAIL PROTECTED]>
To: "Inércia Sensorial" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Saturday, July 07, 2001 8:50 PM
Subject: Re: [PHP] require_once(); questions
> Actually this does work:
> //First_File.php
> <?
> $test = "I am the test variable";
> ?>
>
> //Second_File.php
> <?
> echo $test;
> ?>
>
> //Active file
> <?
> require "First_File.php";
> require "Second_File.php";
> ?>
>
> That would produce the output of :
> I am the test variable
>
> An include or require just places the files' contents at the line you
> include it. So after the includes(requires) my parsed code was:
> <?
> $test = "I am the test variable";
> echo $test;
> ?>
>
> Hope this helps
>
> ----- Original Message -----
> From: "Inércia Sensorial" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Saturday, July 07, 2001 5:59 PM
> Subject: [PHP] require_once(); questions
>
>
> > Hi All,
> >
> > I have a function that includes files based on some SQL queries. On
one
> of
> > the first loaded files, I want to define a function and use it on
another
> > included file. Something like:
> >
> > includes file:
> > first_file.inc.php
> > This file has: $test = "Show me!";
> >
> > Then include file:
> > second_file.php
> > This file has: echo $test;
> >
> > But doesn't show nothing, so I guess it is not possible to do. Am I
> right?
> > If so, what's the best alternative?
> >
> > Also, another question.... since it is a function that includes the
> files,
> > the contents of these files are not available outside the function
scope..
> > what's the best way to use it outside the function?
> >
> > Thanks a lot....
> >
> > --
> >
> >
> > Julio Nobrega.
> >
> > A hora está chegando:
> > http://sourceforge.net/projects/toca
> >
> >
> >
> >
> > --
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> >
>
>
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