Re: listing root directory

Greetings, Dan Gelder.
In reply to Your message dated Sunday, January 27, 2008, 05:08:15,

no, that just turns
/home/.machine/user/site.com/work/january/phpstuff/includer.php
into
/home/.machine/user/site.com/work/january/phpstuff/

and you can read my post to understand why that brings me no closer
than before.

What I need to know, I guess, is a way to get the official root folder
so I have
/home/.machine/user/site.com/

And then I can match the strings?? Can't say how to solve it yet.

Wouldn't $_SERVER["SCRIPT_NAME"] do what you are looking for? Don't use
$_SERVER["PHP_SELF"] because that can contain so-called "additional path
information" that users could be adding to the script's URL in some
circumstances (/january/phpstuff/includer.php/extra/stuff) and which,
incidentally, makes $_SERVER["PHP_SELF"] a value that should not be
trusted. Mind you, perhaps it's better to regard all of $SERVER as
untrusted just to err on the side of caution.

Sebastian Lisken

I know not to trust $_SERVER too far, but it doesn't matter, because I
am interested in finding the *included file*'s location, not the base
script. IE:

------
------ folder1/file1.php:
------

echo "Here is an image from another folder:";
include '/folder2/file2.php';

------
------ folder2/file2.php:
------

function getGlobalPrefix()
{
//this is what I need to write:
//it takes the data in __FILE__ and somehow turns that into 'http://
mySite.com/folder2/'
}
echo "<img src='" . getGlobalPrefix() . "/myGreatImage.gif'>";

------

OK, does it make sense now?? :-)

You has an advice but don't get it right.

$dir = dirname(__FILE__); // current script location
if(stripos($_SERVER['DOCUMENT_ROOT'], $dir) === 0)
{
$dir = substr($dir, strlen($_SERVER['DOCUMENT_ROOT']));
echo("<img src=\"{$dir}/image_in_the_same_location_as_script\"/>");
}


--
Sincerely Yours, AnrDaemon <anrdaemon@xxxxxxxxxxx>

.

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