PHP variables not being passed to functions with include/require

Question

I'm trying to make an easy function for including file, but for some reason, the variables aren't being passed to.

In the file test.php:

<?php
function parse($template) {
    require $template;
}

$test = "test";
parse("stuff.php");
?>

And the file stuff.php:

<?php
echo $test;
?>

And I'll get the error:

Notice: Undefined variable: test in /opt/lampp/htdocs/testinggrounds/stuff.php on line 2

However, if I replace this statement in test.php

parse("stuff.php");

with

require "stuff.php";

And it'll work just fine.

wilson · Accepted Answer · 2015-08-03 15:43:50Z

1

Why make a function when she exist ?

You have just to call require 'stuff.php'; and that's all

answered Aug 3, 2015 at 15:43

wilson

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1 Comment

JayJay Over a year ago

I know. I wanted to put $path and $ext inside the function so all I need was to call parse("script") and it'll include path/script.php. If there's really no other way, then I'll do what you said earlier. But thanks anyway! :)

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