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://22.198.133.16:802         


        

Answer 1

A different approach that takes nothing else into account except for the actual file extension from a url:

def fileExt( url ):
    # compile regular expressions
    reQuery = re.compile( r'\?.*$', re.IGNORECASE )
    rePort = re.compile( r':[0-9]+', re.IGNORECASE )
    reExt = re.compile( r'(\.[A-Za-z0-9]+$)', re.IGNORECASE )

    # remove query string
    url = reQuery.sub( "", url )

    # remove port
    url = rePort.sub( "", url )

    # extract extension
    matches = reExt.search( url )
    if None != matches:
        return matches.group( 1 )
    return None

edit: added handling of explicit ports from :1234