如何使用python获取给定url的原始html文本

Question

我在python中使用html2text通过获取任何URL来获取HTML页面的原始文本（包括标记），但我得到了一个错误。

我的代码-

import html2text
import urllib2

proxy = urllib2.ProxyHandler({'http': 'http://<proxy>:<pass>@<ip>:<port>'})
auth = urllib2.HTTPBasicAuthHandler()
opener = urllib2.build_opener(proxy, auth, urllib2.HTTPHandler)
urllib2.install_opener(opener)
html = urllib2.urlopen("http://www.ndtv.com/india-news/this-stunt-for-a-facebook-like-got-the-hyderabad-youth-arrested-740851").read()
print html2text.html2text(html)

错误-

Traceback (most recent call last):
  File "t.py", line 8, in <module>
    html = urllib2.urlopen("http://www.ndtv.com/india-news/this-stunt-for-a-facebook-like-got-the-hyderabad-youth-arrested-740851").read()
  File "/usr/lib/python2.7/urllib2.py", line 127, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.7/urllib2.py", line 404, in open
    response = self._open(req, data)
  File "/usr/lib/python2.7/urllib2.py", line 422, in _open
    '_open', req)
  File "/usr/lib/python2.7/urllib2.py", line 382, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.7/urllib2.py", line 1214, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.7/urllib2.py", line 1184, in do_open
    raise URLError(err)
urllib2.URLError: <urlopen error [Errno 110] Connection timed out>

有人能解释我做错了什么吗？

Answer 1

如果您不需要SSL，那么Python 2.7.x中的这个脚本应该可以工作：

import urllib
url = "http://stackoverflow.com"
f = urllib.urlopen(url)
print f.read()

在Python 3.x中使用urllib.request，而不是urllib

因为urllib2对于Python 2，在Python 3中它被合并到urllib。

http://是必需的。