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

regexp to get the URL path without the file.

url = 'http://domain/dir1/dir2/somefile' url.scan(/^(http://[^/]+)((?:/[^/]+)+(?=/))?/?(?:[^/]+)?$/i).to_s

It can be useful for adding a relative path to this url.