urllib.urlopen 有效，但 urllib2.urlopen 无效 [英] urllib.urlopen works but urllib2.urlopen doesn't

问题描述

我有一个正在测试的简单网站.它在本地主机上运行，​​我可以在我的网络浏览器中访问它.索引页只是运行"这个词.urllib.urlopen 将成功读取页面，但 urllib2.urlopen 不会.这是一个演示问题的脚本(这是实际脚本，而不是其他测试脚本的简化):

I have a simple website I'm testing. It's running on localhost and I can access it in my web browser. The index page is simply the word "running". urllib.urlopen will successfully read the page but urllib2.urlopen will not. Here's a script which demonstrates the problem (this is the actual script and not a simplification of a different test script):

import urllib, urllib2
print urllib.urlopen("http://127.0.0.1").read()  # prints "running"
print urllib2.urlopen("http://127.0.0.1").read() # throws an exception

这是堆栈跟踪:

Traceback (most recent call last):
  File "urltest.py", line 5, in <module>
    print urllib2.urlopen("http://127.0.0.1").read()
  File "C:\Python25\lib\urllib2.py", line 121, in urlopen
    return _opener.open(url, data)
  File "C:\Python25\lib\urllib2.py", line 380, in open
    response = meth(req, response)
  File "C:\Python25\lib\urllib2.py", line 491, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python25\lib\urllib2.py", line 412, in error
    result = self._call_chain(*args)
  File "C:\Python25\lib\urllib2.py", line 353, in _call_chain
    result = func(*args)
  File "C:\Python25\lib\urllib2.py", line 575, in http_error_302
    return self.parent.open(new)
  File "C:\Python25\lib\urllib2.py", line 380, in open
    response = meth(req, response)
  File "C:\Python25\lib\urllib2.py", line 491, in http_response
    'http', request, response, code, msg, hdrs)
  File "C:\Python25\lib\urllib2.py", line 418, in error
    return self._call_chain(*args)
  File "C:\Python25\lib\urllib2.py", line 353, in _call_chain
    result = func(*args)
  File "C:\Python25\lib\urllib2.py", line 499, in http_error_default
    raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 504: Gateway Timeout

有什么想法吗?我可能最终需要 urllib2 的一些更高级的功能，所以我不想只使用 urllib，而且我想了解这个问题.

Any ideas? I might end up needing some of the more advanced features of urllib2, so I don't want to just resort to using urllib, plus I want to understand this problem.

推荐答案

听起来像是您定义了 urllib2 正在接受的代理设置.当它尝试代理127.0.0.01/"时，代理放弃并返回504错误.

Sounds like you have proxy settings defined that urllib2 is picking up on. When it tries to proxy "127.0.0.01/", the proxy gives up and returns a 504 error.

来自晦涩的python urllib2代理问题:

proxy_support = urllib2.ProxyHandler({})
opener = urllib2.build_opener(proxy_support)
print opener.open("http://127.0.0.1").read()

# Optional - makes this opener default for urlopen etc.
urllib2.install_opener(opener)
print urllib2.urlopen("http://127.0.0.1").read()

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