你应该能够做到urlparse（docs：python2，python3）：
from urllib.parse import urlparse# from urlparse import urlparse # Python 2parsed_uri = urlparse('http://stackoverflow.com/questions/1234567/blah-blah-blah-blah' )result = '{uri.scheme}://{uri.netloc}/'.format(uri=parsed_uri)print(result)# gives'http://stackoverflow.com/'
