CODEWITHSUNDEEP
pythonurl
Rikke Bendlin Gammelmark
Rikke Bendlin Gammelmark

Reputation: 970

Changing hostname in a url

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.

Upvotes: 72

Views: 40819

Answers (7)

Nigel Tufnel
Nigel Tufnel

Reputation: 11514

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))

Upvotes: 124

Facundo Batista
Facundo Batista

Reputation: 41

You can always do this trick:

>>> p = parse.urlparse("https://stackoverflow.com/questions/21628852/changing-hostname-in-a-url")
>>> parse.ParseResult(**dict(p._asdict(), netloc='perrito.com.ar')).geturl()
'https://perrito.com.ar/questions/21628852/changing-hostname-in-a-url'

Upvotes: 4

eLRuLL
eLRuLL

Reputation: 18799

I would recommend also using urlsplit and urlunsplit like @linkyndy's answer, but for Python3 it would be:

>>> from urllib.parse import urlsplit, urlunsplit
>>> url = list(urlsplit('https://www.google.dk:80/barbaz'))
>>> url
['https', 'www.google.dk:80', '/barbaz', '', '']
>>> url[1] = 'www.foo.dk:80'
>>> new_url = urlunsplit(url)
>>> new_url
'https://www.foo.dk:80/barbaz'

Upvotes: 6

David Morley
David Morley

Reputation: 81

A simple string replace of the host in the netloc also works in most cases:

>>> p = urlparse.urlparse('https://www.google.dk:80/barbaz')
>>> p._replace(netloc=p.netloc.replace(p.hostname, 'www.foo.dk')).geturl()
'https://www.foo.dk:80/barbaz'

This will not work if, by some chance, the user name or password matches the hostname. You cannot limit str.replace to replace the last occurrence only, so instead we can use split and join:

>>> p = urlparse.urlparse('https://www.google.dk:[email protected]:80/barbaz')
>>> new_netloc = 'www.foo.dk'.join(p.netloc.rsplit(p.hostname, 1))
>>> p._replace(netloc=new_netloc).geturl()
'https://www.google.dk:[email protected]:80/barbaz'

Upvotes: 7

Omid Raha
Omid Raha

Reputation: 10680

Using urlparse and urlunparse methods of urlparse module:

import urlparse

old_url = 'https://www.google.dk:80/barbaz'
url_lst = list(urlparse.urlparse(old_url))
# Now url_lst is ['https', 'www.google.dk:80', '/barbaz', '', '', '']
url_lst[1] = 'www.foo.dk:80'
# Now url_lst is ['https', 'www.foo.dk:80', '/barbaz', '', '', '']
new_url = urlparse.urlunparse(url_lst)

print(old_url)
print(new_url)

Output:

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

Upvotes: 10

Alfe
Alfe

Reputation: 59416

To just replace the host without touching the port in use (if any), use this:

import re, urlparse

p = list(urlparse.urlsplit('https://www.google.dk:80/barbaz'))
p[1] = re.sub('^[^:]*', 'www.foo.dk', p[1])
print urlparse.urlunsplit(p)

prints

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

If you've not given any port, this works fine as well.

If you prefer the _replace way Nigel pointed out, you can use this instead:

p = urlparse.urlsplit('https://www.google.dk:80/barbaz')
p = p._replace(netloc=re.sub('^[^:]*', 'www.foo.dk', p.netloc))
print urlparse.urlunsplit(p)

Upvotes: 3

linkyndy
linkyndy

Reputation: 17900

You can take advantage of urlsplit and urlunsplit from Python's urlparse:

>>> from urlparse import urlsplit, urlunsplit
>>> url = list(urlsplit('https://www.google.dk:80/barbaz'))
>>> url
['https', 'www.google.dk:80', '/barbaz', '', '']
>>> url[1] = 'www.foo.dk:80'
>>> new_url = urlunsplit(url)
>>> new_url
'https://www.foo.dk:80/barbaz'

As the docs state, the argument passed to urlunsplit() "can be any five-item iterable", so the above code works as expected.

Upvotes: 28

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