urllib2 为 url 抛出错误，而它在浏览器中正确打开 [英] urllib2 is throwing an error for an url , while it's opening properly in browser

问题描述

我正在尝试像这样通过python打开一个网址

I am trying to open an url through python like this

  import urllib2
  f = urllib2.urlopen('http://www.futurebazaar.com/Search/laptop')

它抛出以下错误

文件C:\Python26\lib\urllib2.py"，第 1134 行，在 do_open 中r = h.getresponse() 文件C:\Python26\lib\httplib.py"，行986，在得到响应response.begin() 文件C:\Python26\lib\httplib.py"，行391，在开始版本、状态、原因 = self._read_status() 文件"C:\Python26\lib\httplib.py", 行355，在_read_status引发 BadStatusLine(line) httplib.BadStatusLine

File "C:\Python26\lib\urllib2.py", line 1134, in do_open r = h.getresponse() File "C:\Python26\lib\httplib.py", line 986, in getresponse response.begin() File "C:\Python26\lib\httplib.py", line 391, in begin version, status, reason = self._read_status() File "C:\Python26\lib\httplib.py", line 355, in _read_status raise BadStatusLine(line) httplib.BadStatusLine

但是这个网址是通过浏览器打开的.

But this url is opening via browser.

推荐答案

网站坏了.如果未提供可选的接受"标头，站点将关闭连接而不响应；这是无效行为.

The website is broken. If the optional "Accept" header isn't supplied, the site closes the connection without responding; this is invalid behavior.

解决方法:

import urllib2
req = urllib2.Request('http://www.futurebazaar.com/Search/laptop')
req.add_header('Accept', '*/*')
f = urllib2.urlopen(req)

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