Ok how do i use regex to remove http AND/OR www just to get http://www.domain.com/ into domain.com
Assume x as any kind of TLD or cTLD
Input example:
www.domain.x
Output:
domain.x
3 Answers
Don't use regex, use urlparse to get netloc
>>> x = 'http://www.domain.com/'
>>> from urlparse import urlparse
>>> o = urlparse(x)
>>> o
ParseResult(scheme='http', netloc='www.domain.com', path='/', params='', query='', fragment='')
>>>
and then
>>> o.netloc
'www.domain.com'
>>> if o.netloc.startswith('www.'): print o.netloc[4:]
...
domain.com
>>>
3 Comments
Janne Karila
o.netloc.startswith('www.') would be more appropriate than 'www' in o.netloc
pyfunc
@Janne Karila: Thanks Janne. Lost that completely in quick answering. Thats ofcourse the correct way and not the one I presented. It is infact incorrect.
firephil
python 3.5 : from urllib.parse import urlparse
If you really want to use regular expressions instead of urlparse() or splitting the string:
>>> domain = 'http://www.example.com/'
>>> re.match(r'(?:\w*://)?(?:.*\.)?([a-zA-Z-1-9]*\.[a-zA-Z]{1,}).*', domain).groups()[0]
example.com
The regular expression might a bit simplistic, but works. It's also not replacing, but I think getting the domain out is easier.
To support domains like 'co.uk', one can do the following:
>>> p = re.compile(r'(?:\w*://)?(?:.*?\.)?(?:([a-zA-Z-1-9]*)\.)?([a-zA-Z-1-9]*\.[a-zA-Z]{1,}).*')
>>> p.match(domain).groups()
('google', 'co.uk')
So you got to check the result for domains like 'co.uk', and join the result again in such a case. Normal domains should work OK. I could not make it work when you have multiple subdomains.
One-liner without regular expressions or fancy modules:
>>> domain = 'http://www.example.com/'
>>> '.'.join(domain.replace('http://','').split('/')[0].split('.')[-2:])
3 Comments
geertjanvdk
I managed to paste the wrong regex in my initial post, but it's now edited with the correct one.
geertjanvdk
@Natsume Made me think, and I've updated the regex so 'http://' is optional and it accept any protocol, like 'https://' or 'bzr://'.
geertjanvdk
@Wooble I wouldn't say horribly, since it returns 'co.uk', but I understand the problem. I'm adding a solution for this.
Here is one of the way to do it:
>>>import re
>>>str1 = 'http://www.domain.x/'
>>>p1 = re.compile('http://www.|/')
>>>out = p1.sub('',str1)
2 Comments
geertjanvdk
Nice, but it does not cover where 'www' would be missing from the URL.
user1242393
one can use match from re as below to check if required substring 'www' exists or not : >>> print p1.match("www")
>>> ' spacious '.lstrip()'spacious '>>> 'www.example.com'.lstrip('cmowz.')'example.com'