How to safely get the file extension from a URL?

Question

Consider the following URLs

 http://m3u.com/tunein.m3u http://asxsomeurl.com/listen.asx:8024 http://www.plssomeotherurl.com/station.pls?id=111 http://22.198.133.16:8024 

Whats the proper way to determine the file extensions (.m3u/.asx/.pls)? Obviously the last one doesn't have a file extension.

EDIT: I forgot to mention that m3u/asx/pls are playlists (textfiles) for audio streams and must be parsed differently. The goal determine the extension and then send the url to the proper parsing-function. E.g.

 url = argv[1] ext = GetExtension(url) if ext == "pls":   realurl = ParsePLS(url) elif ext == "asx":   realurl = ParseASX(url) (etc.) else:   realurl = url Play(realurl) 

GetExtension() should return the file extension (if any), preferrably without connecting to the URL.

Answer 1

Use urlparse to parse the path out of the URL, then os.path.splitext to get the extension.

import urlparse, os  url = 'http://www.plssomeotherurl.com/station.pls?id=111' path = urlparse.urlparse(url).path ext = os.path.splitext(path)[1] 

Note that the extension may not be a reliable indicator of the type of the file. The HTTP Content-Type header may be better.