Print file name of including file, not included

Question

I have example1.php and example2.php that include example3.php. In example3.php I have:

<link rel="canonical" href="http://website.com/<?php echo basename(__FILE__); ?>">

What I wanted is to have example1.php/example2.php in href dependent on the file including example3.php. The problem is that of course __FILE__ outputs example3.php. Is it possible to get the filename of the including file, not the included one?

Please don't suggesting using the XSS vulnerable SERVER['HTTP_HOST'].$_SERVER['REQUEST_URI'].

You could set a variable before "including", then reference that variable. A pain, but it would work. — Cole Tobin, Apr 02 '13 at 01:12

Jhonathan H. · Answer 1 · 2013-04-02T00:37:04.053

1

IF you dont want to use any $_SERVER[''] types to get the file name You can use

pathinfo(__FILE__, PATHINFO_BASENAME)

to just get the filename you want

edited Apr 02 '13 at 00:37

answered Apr 02 '13 at 00:30

Jhonathan H.

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Oh, thanks. And `pathinfo(__FILE__, PATHINFO_BASENAME)` is not subject to possible XSS vulnerabilities, right? :) – MultiformeIngegno Apr 02 '13 at 00:38
yeah, sorry it's not that doesn't output nothing, it outputs still example3.php – MultiformeIngegno Apr 02 '13 at 00:43
yes thats what it does.. thats what you intended to get right? – Jhonathan H. Apr 02 '13 at 00:45
No. I wanted to print example1.php or example2.php according to which one of the 2 files was including example3.php :) – MultiformeIngegno Apr 02 '13 at 00:47
i dont get it. can u further illustrate what you suppose trying to do? :) – Jhonathan H. Apr 02 '13 at 00:50
This is what I need: example1.php and example2.php include example3.php which contains `` – MultiformeIngegno Apr 02 '13 at 00:55

score 1 · Accepted Answer · answered Apr 02 '13 at 01:09

(Warning: This trick may not scale, but will work in this exact case.)

Use element 0 of the return from get_included_files(). That will be the original file that started the chain

inside example3.php:

$gif = get_included_files();

then you can do

<link rel="canonical" href="http://website.com/<?php echo basename($gif[0]); ?>">

Print file name of including file, not included

2 Answers2