python - Changing hostname in a url

Question

I am trying to use python to change the hostname in a url, and have been playing around with the urlparse module for a while now without finding a satisfactory solution. As an example, consider the url:

https://www.google.dk:80/barbaz

I would like to replace "www.google.dk" with e.g. "www.foo.dk", so I get the following url:

https://www.foo.dk:80/barbaz.

So the part I want to replace is what urlparse.urlsplit refers to as hostname. I had hoped that the result of urlsplit would let me make changes, but the resulting type ParseResult doesn't allow me to. If nothing else I can of course reconstruct the new url by appending all the parts together with +, but this would leave me with some quite ugly code with a lot of conditionals to get "://" and ":" in the correct places.

Answer 1

You can use urllib.parse.urlparse function and ParseResult._replace method (Python 3):

>>> import urllib.parse
>>> parsed = urllib.parse.urlparse("https://www.google.dk:80/barbaz")
>>> replaced = parsed._replace(netloc="www.foo.dk:80")
>>> print(replaced)
ParseResult(scheme='https', netloc='www.foo.dk:80', path='/barbaz', params='', query='', fragment='')

If you're using Python 2, then replace urllib.parse with urlparse.

ParseResult is a subclass of namedtuple and _replace is a namedtuple method that:

returns a new instance of the named tuple replacing specified fields with new values

UPDATE:

As @2rs2ts said in the comment netloc attribute includes a port number.

Good news: ParseResult has hostname and port attributes. Bad news: hostname and port are not the members of namedtuple, they're dynamic properties and you can't do parsed._replace(hostname="www.foo.dk"). It'll throw an exception.

If you don't want to split on : and your url always has a port number and doesn't have username and password (that's urls like "https://username:[email protected]:80/barbaz") you can do:

parsed._replace(netloc="{}:{}".format(parsed.hostname, parsed.port))