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Python urllib2 cannot open localhost on alternate port (not 80)? Error 10013

开发者 https://www.devze.com 2023-03-22 18:02 出处:网络
Here is my server.py: import BaseHTTPServer import SocketServer class TestRequestHandler(BaseHTTPServer.BaseHTTPRequestHandler):

Here is my server.py:

import BaseHTTPServer
import SocketServer

class TestRequestHandler(BaseHTTPServer.BaseHTTPRequestHandler):
    def do_GET(self):
        self.wfile.write("hello world at %s" % __file__)

server = BaseHTTPServer.HTTPServer(('', 10000), TestRequestHandler)
#server = SocketServer.ThreadingTCPServer(('', 8888), TestRequestHandler)
server.serve_forever()

Here is my client.py:

import urllib2
req = urllib2.Request('http://127.0.0.1:10000/')
handle = urllib2.urlopen(req)
content = handle.read()

Then I start server.py, it works.

When I start client.py, I get this error on Windows 7, Python 2.6:

Traceback (most recent call last):
  File "D:\Dropbox\Forge\urllib-error\client.py", line 3, in <module>
    handle = urllib2.urlopen(req)
  File "C:\Python26\lib\urllib2.py", line 126, in urlopen
    return _opener.open(url, data, timeout)
  File "C:\Python26\lib\urllib2.py", line 391, in open
    response = self._open(req, data)
  File "C:\Python26\lib\urllib2.py", line 409, in _open
    '_open', req)
  File "C:\Python26\lib\urllib2.py", line 369, in _call_chain
    result = func(*args)
  File "C:\Python26\lib\urllib2.py", line 1161, in http_open
    return self.do_open(httplib.HTTPConnection, req)
  File "C:\Python26\lib\urllib2.py", line 1136, in do_open
    raise URLError(err)
urllib2.URLError: <urlopen error [Errno 10013] An attempt was made to access a socket in a way forbidden by its access permissions>

When I open http://127.0.0.1:10000/ from a browser,开发者_Python百科 it works. It means server is fine.

When I replace http://127.0.0.1:10000/ with http://127.0.0.1/ or http://127.0.0.1:80/ in client.py, everything works fine (this is another web server on port 80 - apache).

What do I do wrong?

UPD: Same error when I use this client2.py:

import urllib2
handle = urllib2.urlopen('http://127.0.0.1:10000/')
content = handle.read()

UPD: Problem solved. It was my firewall. When disabled, everything works. Stupid, stupid me. Thanks for reading :-)


The local firewall prevented the connection. When it's disabled, everything works.

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