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

Propose a much more readable solution (in Python, but applies to any regex):

def url_path_to_dict(path):
    pattern = (r'^'
               r'((?P.+?)://)?'
               r'((?P.+?)(:(?P.*?))?@)?'
               r'(?P.*?)'
               r'(:(?P\d+?))?'
               r'(?P/.*?)?'
               r'(?P[?].*?)?'
               r'$'
               )
    regex = re.compile(pattern)
    m = regex.match(path)
    d = m.groupdict() if m is not None else None

    return d

def main():
    print url_path_to_dict('http://example.example.com/example/example/example.html')

Prints:

{
'host': 'example.example.com', 
'user': None, 
'path': '/example/example/example.html', 
'query': None, 
'password': None, 
'port': None, 
'schema': 'http'
}