从URL获取协议+主机名 [英] Get protocol + host name from URL

问题描述

在我的Django应用中，我需要在 request.META.get（'HTTP_REFERER'）的引荐来源网址中获取主机名及其协议，以便从URL像这样：

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.com/whatever/blah/blah/?v1=0&v2=blah+blah ...

• 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.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/

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

我查看了其他相关问题并发现了有关urlparse的信息，

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'

推荐答案

您应该可以使用 urlparse （文档： python2 ， python3 ）：

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/'

这篇关于从URL获取协议+主机名的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！