Extract domain from URL in python [duplicate]

Answer 1

Instead of regex or hand-written solutions, you can use python's urlparse

from urllib.parse import urlparse

print (urlparse('http://abc.hostname.com/somethings/anything/'))
>> ParseResult(scheme='http', netloc='abc.hostname.com', path='/somethings/anything/', params='', query='', fragment='')

print (urlparse('http://abc.hostname.com/somethings/anything/').netloc)
>> abc.hostname.com

To get without the subdomain

t = urlparse('http://abc.hostname.com/somethings/anything/').netloc
print ('.'.join(t.split('.')[1:]))
>> hostname.com

Answer 2

Try this code:

from tldextract import extract

tsd, td, tsu = extract("http://abc.hostname.com/somethings/anything/") # prints abc, hostname, com

url = td + '.' + tsu # will prints as hostname.com

print url

Answer 3

Try:

from urlparse import urlparse

parsed = urlparse('http://abc.hostname.com/somethings/anything/')
domain = parsed.netloc.split(".")[1:]
host = ".".join(domain)
print host  # will prints hostname.com

Answer 4

For parsing the domain of a URL in Python 3, you can use:

from urllib.parse import urlparse

domain = urlparse('http://www.example.test/foo/bar').netloc
print(domain) # --> www.example.test

However, for reliably parsing the top-level domain (example.test in this example), you need to install a specialized library (e.g., tldextract).

Answer 5

Assuming you have it in an accessible string, and assuming we want to be generic for having multiple levels on the top domain, you could:

token=my_string.split('http://')[1].split('/')[0]
top_level=token.split('.')[-2]+'.'+token.split('.')[-1]

We split first by the http:// to remove that from the string. Then we split by the / to remove all directory or sub-directory parts of the string, and then the [-2] means we take the second last token after a ., and append it with the last token, to give us the top level domain.

There are probably more graceful and robust ways to do this, for example if your website is http://.com it will break, but its a start :)

Answer 6

best way i found is:

from six.moves.urllib.parse import urlparse

t = urlparse('http://asas.abc.hostname.com/somethings/anything/').netloc

print('.'.join(t.split('.')[-2:]))