[php] Get parent directory of running script

In PHP, what would be the cleanest way to get the parent directory of the current running script relative to the www root? Assume I have:

$_SERVER['SCRIPT_NAME'] == '/relative/path/to/script/index.php'

Or just:

$something_else == '/relative/path/to/script/'

And I need to get /relative/path/to/ with slashes properly inserted. What would you suggest? A one liner is preferred.

EDIT

I need to get a path relative to the www root, dirname(__FILE__) gives me an absolute path in the filesystem so that won't work. $_SERVER['SCRIPT_NAME'] on the other hand 'starts' at the www root.

This is a function that I use. Created it once so I always have this functionality:

function getDir(){
    $directory = dirname(__FILE__);
    $directory = explode("/",$directory);
    $findTarget = 0;
    $targetPath = "";
    foreach($directory as $dir){
        if($findTarget == 1){
            $targetPath = "".$targetPath."/".$dir."";
        }
        if($dir == "public_html"){
            $findTarget = 1;
        }
    }
    return "http://www.".$_SERVER['SERVER_NAME']."".$targetPath."";
}

Try this. Works on both windows or linux server..

str_replace('\\','/',dirname(dirname(__FILE__)))

If I properly understood your question, supposing your running script is

/relative/path/to/script/index.php

This would give you the parent directory of your running script relative to the document www:

$parent_dir = dirname(dirname($_SERVER['SCRIPT_NAME'])) . '/';
//$parent_dir will be '/relative/path/to/'

If you want the parent directory of your running script relative to server root:

$parent_dir = dirname(dirname($_SERVER['SCRIPT_FILENAME'])) . '/';
//$parent_dir will be '/root/some/path/relative/path/to/'

$dir = dirname($file) . DIRECTORY_SEPARATOR;

I Hope this will help you.

echo getcwd().'<br>'; // getcwd() will return current working directory
echo dirname(getcwd(),1).'<br>';
echo dirname(getcwd(),2).'<br>';
echo dirname(getcwd(),3).'<br>';

Output :

C:\wamp64\www\public_html\step
C:\wamp64\www\public_html
C:\wamp64\www
C:\wamp64

Here is what I use since I am not running > 5.2

function getCurrentOrParentDirectory($type='current')
{
    if ($type == 'current') {
        $path = dirname(__FILE__);  
    } else {
        $path = dirname(dirname(__FILE__));
    }
    $position = strrpos($path, '/') + 1;
    return substr($path, $position);
}

Double dirname with file as suggested by @mike b for the parent directory, and current directory is found by just using that syntax once.

Note this function only returns the NAME, slashes have to be added afterwards.

Got it myself, it's a bit kludgy but it works:

substr(dirname($_SERVER['SCRIPT_NAME']), 0, strrpos(dirname($_SERVER['SCRIPT_NAME']), '/') + 1)

So if I have /path/to/folder/index.php, this results in /path/to/.

I hope this will help

function get_directory(){
    $s = empty($_SERVER["HTTPS"]) ? '' : ($_SERVER["HTTPS"] == "on") ? "s" : "";
    $protocol = substr(strtolower($_SERVER["SERVER_PROTOCOL"]), 0, strpos(strtolower($_SERVER["SERVER_PROTOCOL"]), "/")) . $s;
    $port = ($_SERVER["SERVER_PORT"] == "80") ? "" : (":".$_SERVER["SERVER_PORT"]);
return $protocol . "://" . $_SERVER['SERVER_NAME'] . dirname($_SERVER['PHP_SELF']);
}
define("ROOT_PATH", get_directory()."/" );
echo ROOT_PATH;

As of PHP 5.3.0 you can use __DIR__ for this purpose.

The directory of the file. If used inside an include, the directory of the included file is returned. This is equivalent to dirname(__ FILE__).

See PHP Magic constants.

C:\www>php --version
PHP 5.5.6 (cli) (built: Nov 12 2013 11:33:44)
Copyright (c) 1997-2013 The PHP Group
Zend Engine v2.5.0, Copyright (c) 1998-2013 Zend Technologies

C:\www>php -r "echo __DIR__;"
C:\www

This is also a possible solution

$relative = '/relative/path/to/script/';
$absolute = __DIR__. '/../' .$relative;

To get the parentdir of the current script.

$parent_dir = dirname(__DIR__);

Fugly, but this will do it:

substr($_SERVER['SCRIPT_NAME'], 0, strpos($_SERVER['SCRIPT_NAME'],basename($_SERVER['SCRIPT_NAME'])))