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:
I should get:
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'
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)
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
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 ...
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'
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
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'
if you think your url is valid then this will work all the time
domain = "http://google.com".split("://")[1].split("/")[0]
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'
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.
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)
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
>>> import urlparse
>>> url = 'http://stackoverflow.com/questions/1234567/blah-blah-blah-blah'
>>> urlparse.urljoin(url, '/')
'http://stackoverflow.com/'
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
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/
Source: Stackoverflow.com