带有 urllib 模块的 Python 程序

Question

各位

下面的程序用于找出页面http://whatismyipaddress.com/中给出的IP地址

import urllib2
import re

response = urllib2.urlopen('http://whatismyipaddress.com/')

p = response.readlines()
for line in p:
    ip = re.findall(r'(\d+.\d+.\d+.\d+)',line)
    print ip

但我无法解决该问题，因为它给出了以下错误

Traceback (most recent call last):
  File "Test.py", line 5, in <module>
  response = urllib2.urlopen('http://whatismyipaddress.com/')
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 154, in urlopen
  return opener.open(url, data, timeout)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 437, in open
  response = meth(req, response)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 550, in http_response
  'http', request, response, code, msg, hdrs)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 475, in error
  return self._call_chain(*args)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 409, in _call_chain
  result = func(*args)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 558, in http_error_default
  raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)

urllib2.HTTPError:HTTP 错误 403:禁止

有人知道需要进行哪些更改才能消除错误并获得所需的输出吗？

Answer 1

http错误代码403告诉您服务器出于某种原因不想响应您的请求。在本例中，我认为它是您的查询的用户代理(urllib2 使用的默认值)。

您可以更改用户代理:

opener = urllib2.build_opener()
opener.addheaders = [('User-agent', 'Mozilla/5.0')]
response = opener.open('http://www.whatismyipaddress.com/')

然后你的查询就会起作用。

但不能保证这会继续有效。该网站可以决定阻止自动查询。