Getting parts of a URL (Regex)

Question

Given the URL (single line):
http://test.example.com/dir/subdir/file.html

How can I extract the following parts using regular expressions:

1. The Subd

Answer 1

I found the highest voted answer (hometoast's answer) doesn't work perfectly for me. Two problems:

1. It can not handle port number.
2. The hash part is broken.

The following is a modified version:

^((http[s]?|ftp):\/)?\/?([^:\/\s]+)(:([^\/]*))?((\/\w+)*\/)([\w\-\.]+[^#?\s]+)(\?([^#]*))?(#(.*))?$

Position of parts are as follows:

int SCHEMA = 2, DOMAIN = 3, PORT = 5, PATH = 6, FILE = 8, QUERYSTRING = 9, HASH = 12

Edit posted by anon user:

function getFileName(path) {
    return path.match(/^((http[s]?|ftp):\/)?\/?([^:\/\s]+)(:([^\/]*))?((\/[\w\/-]+)*\/)([\w\-\.]+[^#?\s]+)(\?([^#]*))?(#(.*))?$/i)[8];
}