Re: Which file called the function?
| From: | Shawn McKenzie | Date: | Thu, 20 Dec 2007 15:32:07 +0000 |
| Subject: | Re: Which file called the function? | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-266161@lists.php.net to get a copy of this message | ||
This should work:
global.php
<?
function myFunc($file) { echo $file; }
?>
one.php
<?
include( 'global.php' );
echo 'You are in file: ';
myFunc(__FILE__);
?>
two.php
<?
include( 'global.php' );
echo 'You are in file: ';
myFunc(__FILE__);
?>
Christoph Boget wrote:
> Let's say I have the following 3 files
>
> global.php
> <?
> function myFunc() { echo __FILE__; }
> ?>
>
> one.php
> <?
> include( 'global.php' );
> echo 'You are in file: ';
> myFunc();
> ?>
>
> two.php
> <?
> include( 'global.php' );
> echo 'You are in file: ';
> myFunc();
> ?>
>
> In each case, what is echoed out for __FILE__ is global.php. Apart from
> analyzing the debug_backtrace array, is there any way that myFunc() would
> display "one.php" and "two.php" respectively?
>
> thnx,
> Christoph
>