3

这是我的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()

这是我的client.py

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

然后我启动 server.py,它可以工作。

当我启动 client.py 时,我在Windows 7、Python 2.6上收到此错误:

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>

当我从浏览器打开http://127.0.0.1:10000/时,它可以工作。这意味着服务器很好。

当我在 client.py 中将 http://127.0.0.1:10000/ 替换http://127.0.0.1/http://127.0.0.1:80/,一切正常(这是端口 80 上的另一个 Web 服务器- 阿帕奇)。

我做错了什么?

UPD:当我使用这个client2.py时出现同样的错误:

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

UPD:问题解决了。这是我的防火墙。禁用时,一切正常。愚蠢的,愚蠢的我。谢谢阅读 :-)

4

1 回答 1

5

本地防火墙阻止了连接。当它被禁用时,一切正常。

于 2011-07-24T09:59:19.000 回答