Function doesn't use variables defined above it - php

Question

I have a variable named $url in the top of my file:
<?php
$url = "http://myurl.com";
Later in the same file, I have this code:
<?php
$url = "http://myurl.com";
[...]
function errorOut($error, $type = "info", $rel = "/")
{
echo $url;
}
?>
However, that doesn't work, because it says $url isn't a valid variable. I have to do this:
<?php
$url = "http://myurl.com";
[...]
function errorOut($error, $type = "info", $rel = "/")
{
$url = "http://myurl.com";
echo $url;
}
?>
This doesn't make any sense to me because it shouldn't be out of scope because it's a layer above the function. How do I make it use the earlier $url variable?

Answer 1

They are not in the same scope. You have to let PHP know you will be using that global locally. It is preferable to not use a global and instead pass it as a variable though.
<?php
$url = "http://myurl.com";
[...]
function errorOut($error, $type = "info", $rel = "/")
{
global $url;
echo $url;
}
?>
See Variable scope for more information.

Answer 2

I was talking to someone in an IRC about this (he posted on here too before I joined), he said I should use
define("URL", 'http://example.com');
And whenever I reference that variable I should use URL, not $url

Answer 3

by passing the $url in your function:
function errorOut($error, $type = "info", $rel = "/", $url) //<<< here
and also calling it:
errorOut('...','...','...',$url);
NON WORKING EXAMPLE AS SEEN IN YOUR ANSWER
$a = 'test1';
$b = 'test2';
define ('URL','one');
define ('URL', 'two');
test($a,$b);
function test ($a,$b){
echo $a;
echo $b;
echo URL;
}
Won't work, URL will stay at 'one' // Will only work if you never want to change URL