从URL获取协议+主机名

Question

在我的Django应用程序中,我需要从referrer中获取主机名request.META.get('HTTP_REFERER')及其协议,以便从以下URL中获取:

• https://docs.google.com/spreadsheet/ccc?key=blah-blah-blah-blah#gid=1
• /sf/ask/86419721/
• http://www.example.com
• https://www.other-domain.com/whatever/blah/blah/?v1=0&v2=blah+blah ...

我应该得到:

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

我查看了其他相关的问题,发现了关于urlparse,但是从那以后就没办法了

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

Answer 1

您应该可以使用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/'

Answer 2

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

这是urlparse的更详细版本.它会为您检测域和子域.

从他们的文件:

>>> 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 是一个namedtuple,所以访问你想要的部分很简单.

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

Answer 3

Python3使用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/

Answer 4

纯字符串操作:):

>>> 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'

总而言之,伙计们.

Answer 5

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

Answer 6

标准库函数urllib.parse.urlsplit()就是你所需要的。以下是 Python3 的示例：

>>> import urllib.parse
>>> o = urllib.parse.urlsplit('https://user:pass@www.example.com:8080/dir/page.html?q1=test&q2=a2#anchor1')
>>> o.scheme
'https'
>>> o.netloc
'user:pass@www.example.com: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'

Answer 7

如果您认为自己的网址有效，那么它将一直有效

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

Answer 8

纯字符串操作有什么问题吗：

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

如果您更喜欢附加斜杠，请像这样扩展此脚本：

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

这可能可以优化一点......

Answer 9

这是一个稍微改进的版本:

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

产量

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

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