再次urllib.error.HTTPError：HTTP错误400：错误的请求

Wanu 发表于 Dev

万努

！我试图打开通常在浏览器中打开的网页，但是python只是发誓并且不想工作。

import urllib.request, urllib.error
f = urllib.request.urlopen('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphire')

还有另一种方式

import urllib.request, urllib.error
opener=urllib.request.build_opener()
f=opener.open('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphi
re')

这两个选项都给出一种错误类型：

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python34\lib\urllib\request.py", line 461, in open
    response = meth(req, response)
  File "C:\Python34\lib\urllib\request.py", line 571, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python34\lib\urllib\request.py", line 493, in error
    result = self._call_chain(*args)
  File "C:\Python34\lib\urllib\request.py", line 433, in _call_chain
    result = func(*args)
  File "C:\Python34\lib\urllib\request.py", line 676, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "C:\Python34\lib\urllib\request.py", line 461, in open
    response = meth(req, response)
  File "C:\Python34\lib\urllib\request.py", line 571, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python34\lib\urllib\request.py", line 499, in error
    return self._call_chain(*args)
  File "C:\Python34\lib\urllib\request.py", line 433, in _call_chain
    result = func(*args)
  File "C:\Python34\lib\urllib\request.py", line 579, in http_error_default
    raise HTTPError(req.full_url, code, msg, hdrs, fp)
urllib.error.HTTPError: HTTP Error 400: Bad Request

有任何想法吗？

安迪

此URL似乎正在执行用户代理字符串检查。如果我将Firefox中的用户代理字符串调整为Python-urllib/2.7，则Bad Request您看到的将失败。

在使用时urllib，您可以按照本教程调整用户代理

from urllib.request import FancyURLopener

class MyOpener(FancyURLopener):
    version = 'My new User-Agent'   # Set this to a string you want for your user agent

myopener = MyOpener()
page = myopener.open('http://www.booking.com/reviewlist.html?cc1=tr;pagename=sapphire')

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