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Asked: 2026-05-10T18:52:34+00:00

I have a simple website I’m testing. It’s running on localhost and I can

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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

Here’s the stack trace:

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

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.

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1 Answer

  1. 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.

    From Obscure python urllib2 proxy gotcha:

    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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