Get protocol + host name from URL

Question

In my Django app, I need to get the host name from the referrer in request.META.get('HTTP_REFERER') along with its protocol so that from URLs like:

https://docs.google.com/spreadsheet/ccc?key=blah-blah-blah-blah#gid=1
https://stackoverflow.com/questions/1234567/blah-blah-blah-blah
http://www.example.com
https://www.other-domain.example/whatever/blah/blah/?v1=0&v2=blah+blah

I should get:

https://docs.google.com/
https://stackoverflow.com/
http://www.example.com
https://www.other-domain.example/

I looked over other related questions and found about urlparse, but that didn't do the trick since

>>> urlparse(request.META.get('HTTP_REFERER')).hostname
'docs.google.com'

wim · Accepted Answer · 2018-07-22 15:35:14Z

378

You should be able to do it with urlparse (docs: python2, python3):

from urllib.parse import urlparse
# from urlparse import urlparse  # Python 2
parsed_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/'

edited Jul 22, 2018 at 15:35

wim

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answered Mar 8, 2012 at 23:17

kgr

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6 Comments

SingleNegationElimination Over a year ago

this answer adds a / to the third example http://www.domain.com, but I think this might be a shortcoming of the question, not of the answer.

Gerard Over a year ago

@TokenMacGuy: ya, my bad... didn't notice the missing /

starrify Over a year ago

I don't think this is a good solution, as netloc is not domain: try urlparse.urlparse('http://user:[email protected]:8080') and find it gives parts like 'user:pass@' and ':8080'

SparkAndShine Over a year ago

The urlparse module is renamed to urllib.parse in Python 3. So, from urllib.parse import urlparse

Scone Over a year ago

This answers what the author meant to ask, but not what was actually stated. For those looking for domain name and not hostname (as this solution provides) I suggest looking at dm03514's answer that is currently below. Python's urlparse cannot give you domain names. Something that seems an oversight.

|

dm03514 · Accepted Answer · 2015-06-03 12:57:26Z

https://github.com/john-kurkowski/tldextract

This is a more verbose version of urlparse. It detects domains and subdomains for you.

From their documentation:

>>> import tldextract
>>> tldextract.extract('http://forums.news.cnn.com/')
ExtractResult(subdomain='forums.news', domain='cnn', suffix='com')
>>> tldextract.extract('http://forums.bbc.co.uk/') # United Kingdom
ExtractResult(subdomain='forums', domain='bbc', suffix='co.uk')
>>> tldextract.extract('http://www.worldbank.org.kg/') # Kyrgyzstan
ExtractResult(subdomain='www', domain='worldbank', suffix='org.kg')

ExtractResult is a namedtuple, so it's simple to access the parts you want.

>>> ext = tldextract.extract('http://forums.bbc.co.uk')
>>> ext.domain
'bbc'
>>> '.'.join(ext[:2]) # rejoin subdomain and domain
'forums.bbc'

This is the correct answer for the question as written, how to get the DOMAIN name. The chosen solution provides the HOSTNAME, which I believe is what the author wanted in the first place.

Marc SJ · Accepted Answer · 2013-12-22 11:20:29Z

52

Python3 using urlsplit:

from urllib.parse import urlsplit
url = "http://stackoverflow.com/questions/9626535/get-domain-name-from-url"
base_url = "{0.scheme}://{0.netloc}/".format(urlsplit(url))
print(base_url)
# http://stackoverflow.com/

answered Dec 22, 2013 at 11:20

Marc SJ

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Comments

png · Accepted Answer · 2015-08-25 22:02:51Z

35

>>> import urlparse
>>> url = 'http://stackoverflow.com/questions/1234567/blah-blah-blah-blah'
>>> urlparse.urljoin(url, '/')
'http://stackoverflow.com/'

answered Aug 25, 2015 at 22:02

png

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2 Comments

Jeff Bowen Over a year ago

For Python 3 the import is from urllib.parse import urlparse.

Sunny Patel Over a year ago

The argument doesn't seem intuitive, but it works great as a very simple native solution

SebMa · Accepted Answer · 2018-02-05 16:21:40Z

26

Pure string operations :):

>>> url = "http://stackoverflow.com/questions/9626535/get-domain-name-from-url"
>>> url.split("//")[-1].split("/")[0].split('?')[0]
'stackoverflow.com'
>>> url = "stackoverflow.com/questions/9626535/get-domain-name-from-url"
>>> url.split("//")[-1].split("/")[0].split('?')[0]
'stackoverflow.com'
>>> url = "http://foo.bar?haha/whatever"
>>> url.split("//")[-1].split("/")[0].split('?')[0]
'foo.bar'

That's all, folks.

edited Feb 5, 2018 at 16:21

answered Apr 13, 2016 at 21:26

SebMa

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2 Comments

Simon Steinberger Over a year ago

Good and simple option, but fails in some cases, e.g. foo.bar?haha

SebMa Over a year ago

@SimonSteinberger :-) How'bout this : url.split("//")[-1].split("/")[0].split('?')[0] :-))

famzah · Accepted Answer · 2020-03-03 20:38:22Z

The standard library function urllib.parse.urlsplit() is all you need. Here is an example for Python3:

>>> import urllib.parse
>>> o = urllib.parse.urlsplit('https://user:[email protected]:8080/dir/page.html?q1=test&q2=a2#anchor1')
>>> o.scheme
'https'
>>> o.netloc
'user:[email protected]:8080'
>>> o.hostname
'www.example.com'
>>> o.port
8080
>>> o.path
'/dir/page.html'
>>> o.query
'q1=test&q2=a2'
>>> o.fragment
'anchor1'
>>> o.username
'user'
>>> o.password
'pass'

Pedro Lobito · Accepted Answer · 2019-04-13 07:32:22Z

9

if you think your url is valid then this will work all the time

domain = "http://google.com".split("://")[1].split("/")[0]

edited Apr 13, 2019 at 7:32

Pedro Lobito

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answered May 7, 2018 at 12:57

Jeeva

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3 Comments

Pedro Lobito Over a year ago

The last split is wrong, there are no more forward slashes to split.

Jeeva Over a year ago

it's won't be a problem, if there are no more slashes then, the list will return with one element. so it will work whether there is a slash or not

Pedro Lobito Over a year ago

I edited your answer the be able to remove the down-vote. Nice explanation. Tks.

Faiz · Accepted Answer · 2017-11-23 17:03:02Z

Here is a slightly improved version:

urls = [
    "http://stackoverflow.com:8080/some/folder?test=/questions/9626535/get-domain-name-from-url",
    "Stackoverflow.com:8080/some/folder?test=/questions/9626535/get-domain-name-from-url",
    "http://stackoverflow.com/some/folder?test=/questions/9626535/get-domain-name-from-url",
    "https://StackOverflow.com:8080?test=/questions/9626535/get-domain-name-from-url",
    "stackoverflow.com?test=questions&v=get-domain-name-from-url"]
for url in urls:
    spltAr = url.split("://");
    i = (0,1)[len(spltAr)>1];
    dm = spltAr[i].split("?")[0].split('/')[0].split(':')[0].lower();
    print dm

Output

stackoverflow.com
stackoverflow.com
stackoverflow.com
stackoverflow.com
stackoverflow.com

Fiddle: https://pyfiddle.io/fiddle/23e4976e-88d2-4757-993e-532aa41b7bf0/?i=true

IMHO the best solution, because simple and it considers all sorts of rare cases. Thanks!
This is not a solution for the question because you do not provide protocol (https:// or http://)

Simon Steinberger · Accepted Answer · 2013-06-07 22:06:56Z

5

Is there anything wrong with pure string operations:

url = 'http://stackoverflow.com/questions/9626535/get-domain-name-from-url'
parts = url.split('//', 1)
print parts[0]+'//'+parts[1].split('/', 1)[0]
>>> http://stackoverflow.com

If you prefer having a trailing slash appended, extend this script a bit like so:

parts = url.split('//', 1)
base = parts[0]+'//'+parts[1].split('/', 1)[0]
print base + (len(url) > len(base) and url[len(base)]=='/'and'/' or '')

That can probably be optimized a bit ...

answered Jun 7, 2013 at 22:06

Simon Steinberger

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1 Comment

Gerard Over a year ago

it's not wrong but we got a tool that already does the work, let's not reinvent the wheel ;)

Orix Au Yeung · Accepted Answer · 2018-11-02 03:51:41Z

3

I know it's an old question, but I too encountered it today. Solved this with an one-liner:

import re
result = re.sub(r'(.*://)?([^/?]+).*', '\g<1>\g<2>', url)

edited Nov 2, 2018 at 3:51

answered Nov 2, 2018 at 3:32

Orix Au Yeung

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Comments

SingleNegationElimination · Accepted Answer · 2012-03-09 01:43:03Z

2

This is a bit obtuse, but uses urlparse in both directions:

import urlparse
def uri2schemehostname(uri):
    urlparse.urlunparse(urlparse.urlparse(uri)[:2] + ("",) * 4)

that odd ("",) * 4 bit is because urlparse expects a sequence of exactly len(urlparse.ParseResult._fields) = 6

answered Mar 9, 2012 at 1:43

SingleNegationElimination

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Comments

kiwi · Accepted Answer · 2019-11-05 21:49:35Z

2

It could be solved by re.search()

import re
url = 'https://docs.google.com/spreadsheet/ccc?key=blah-blah-blah-blah#gid=1'
result = re.search(r'^http[s]*:\/\/[\w\.]*', url).group()
print(result)

#result
'https://docs.google.com'

answered Nov 5, 2019 at 21:49

kiwi

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1 Comment

biscuit314 Over a year ago

Does not include port

Mirko · Accepted Answer · 2020-09-25 01:52:22Z

2

You can simply use urljoin with relative root '/' as second argument:

import urllib.parse


url = 'https://stackoverflow.com/questions/9626535/get-protocol-host-name-from-url'
root_url = urllib.parse.urljoin(url, '/')
print(root_url)

edited Sep 25, 2020 at 1:52

answered Jul 15, 2020 at 18:47

Mirko

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Comments

Praveen Kumar · Accepted Answer · 2021-12-21 17:42:30Z

2

This is the simple way to get the root URL of any domain.

from urllib.parse import urlparse

url = urlparse('https://stackoverflow.com/questions/9626535/')
root_url = url.scheme + '://' + url.hostname
print(root_url) # https://stackoverflow.com

answered Dec 21, 2021 at 17:42

Praveen Kumar

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Comments

Juraj · Accepted Answer · 2018-06-22 03:21:14Z

-1

If it contains less than 3 slashes thus you've it got and if not then we can find the occurrence between it:

import re

link = http://forum.unisoftdev.com/something

slash_count = len(re.findall("/", link))
print slash_count # output: 3

if slash_count > 2:
   regex = r'\:\/\/(.*?)\/'
   pattern  = re.compile(regex)
   path = re.findall(pattern, url)

   print path

edited Jun 22, 2018 at 3:21

answered Jun 21, 2018 at 20:55

Juraj

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Comments

cieunteung · Accepted Answer · 2019-02-25 18:34:10Z

-1

to get domain/hostname and Origin*

url = 'https://stackoverflow.com/questions/9626535/get-protocol-host-name-from-url'
hostname = url.split('/')[2] # stackoverflow.com
origin = '/'.join(url.split('/')[:3]) # https://stackoverflow.com

*Origin is used in XMLHttpRequest headers

answered Feb 25, 2019 at 18:34

cieunteung

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