Python urllib 和 urllib2 不打开 localhost URL？

Question

在Python中，我可以使用urllib2（还有urllib）来打开像谷歌这样的外部网址。不过，当我尝试打开本地网址时遇到了一些问题。我在端口8280上运行了一个Python的SimpleHTTPServer，我可以通过http://localhost:8280/成功访问它。

python -m SimpleHTTPServer 8280

值得一提的是，我正在使用Ubuntu系统，并且有一个叫做CNTLM的工具在运行，它负责处理我们公司网络代理的认证。因此，wget在本地网址上也无法正常工作，所以我觉得这不是urllib的问题！

测试脚本（test_urllib2.py）：

import urllib2

print "Opening Google..."
google = urllib2.urlopen("http://www.google.com/")
print google.read(100)
print "Google opened."

print "Opening localhost..."
localhost = urllib2.urlopen("http://localhost:8280/")
print localhost.read(100)
print "localhost opened."

输出：

$ ./test_urllib2.py 
Opening Google...
<!doctype html><html><head><meta http-equiv="content-type" content="text/html; charset=ISO-8859-1"><
Google opened.
Opening localhost...
Traceback (most recent call last):
  File "./test_urllib2.py", line 10, in <module>
    localhost = urllib2.urlopen("http://localhost:8280/")
  File "/usr/lib/python2.6/urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "/usr/lib/python2.6/urllib2.py", line 397, in open
    response = meth(req, response)
  File "/usr/lib/python2.6/urllib2.py", line 510, in http_response
    'http', request, response, code, msg, hdrs)
  File "/usr/lib/python2.6/urllib2.py", line 429, in error
    result = self._call_chain(*args)
  File "/usr/lib/python2.6/urllib2.py", line 369, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.6/urllib2.py", line 605, in http_error_302
    return self.parent.open(new, timeout=req.timeout)
  File "/usr/lib/python2.6/urllib2.py", line 391, in open
    response = self._open(req, data)
  File "/usr/lib/python2.6/urllib2.py", line 409, in _open
    '_open', req)
  File "/usr/lib/python2.6/urllib2.py", line 369, in _call_chain
    result = func(*args)
  File "/usr/lib/python2.6/urllib2.py", line 1161, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "/usr/lib/python2.6/urllib2.py", line 1134, in do_open
    r = h.getresponse()
  File "/usr/lib/python2.6/httplib.py", line 986, in getresponse
    response.begin()
  File "/usr/lib/python2.6/httplib.py", line 391, in begin
    version, status, reason = self._read_status()
  File "/usr/lib/python2.6/httplib.py", line 355, in _read_status
    raise BadStatusLine(line)
httplib.BadStatusLine

解决方案：问题确实是因为我在公司网络代理后面使用了CNTLM（具体为什么会造成问题我不太确定）。解决办法是使用ProxyHandler：

proxy_support = urllib2.ProxyHandler({})
opener = urllib2.build_opener(proxy_support)
print opener.open("http://localhost:8380/").read(100)

感谢loki2302给我指引到这里。

ubuntu urllib2 urllib localhost SimpleHTTPServer ProxyHandler cntlm network proxy

Python urllib 和 urllib2 不打开 localhost URL？

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