python - xmlrpc 服务器的问题

Question

我使用 xmlrpc 服务器运行简单示例，然后按键盘上的 Ctrl-C :)。

from SimpleXMLRPCServer import SimpleXMLRPCServerfrom time import sleepimport threading,timeclass Test(threading.Thread):    def __init__(self):        threading.Thread.__init__(self)        self.test1 = 0    def test(self):        return self.test1    def run(self):        while(1):            time.sleep(1)            self.test1 = self.test1 + 1ts = Test()ts.start()server =  SimpleXMLRPCServer(("localhost",8888))server.register_instance(ts)server.serve_forever()

按下键盘后出错:

  File "/usr/lib/python2.7/SocketServer.py", line 225, in serve_forever    r, w, e = select.select([self], [], [], poll_interval)KeyboardInterrupt

客户端

 from xmlrpclib import ServerProxyr=ServerProxy("http://localhost:8888")print r.test() 

等待连接，没有错误或警告。在这种情况下如何断开连接？也许这个例子不正确？

Answer 1

使用超时:

Set timeout for xmlrpclib.ServerProxy

编辑

此处链接的答案与 Python 2.7 不兼容。这是修改后的有效代码(在 W7/ActivePython 2.7 上测试):

import xmlrpclib
import httplib

class TimeoutHTTPConnection(httplib.HTTPConnection):

    def __init__(self,host,timeout=10):
        httplib.HTTPConnection.__init__(self,host,timeout=timeout)
        self.set_debuglevel(99)
        #self.sock.settimeout(timeout)

"""
class TimeoutHTTP(httplib.HTTP):
    _connection_class = TimeoutHTTPConnection
    def set_timeout(self, timeout):
        self._conn.timeout = timeout
"""

class TimeoutTransport(xmlrpclib.Transport):
    def __init__(self, timeout=10, *l, **kw):
        xmlrpclib.Transport.__init__(self,*l,**kw)
        self.timeout=timeout

    def make_connection(self, host):
        conn = TimeoutHTTPConnection(host,self.timeout)
        return conn

class TimeoutServerProxy(xmlrpclib.ServerProxy):
    def __init__(self,uri,timeout=10,*l,**kw):
        kw['transport']=TimeoutTransport(timeout=timeout, use_datetime=kw.get('use_datetime',0))
        xmlrpclib.ServerProxy.__init__(self,uri,*l,**kw)

if __name__ == "__main__":
    s=TimeoutServerProxy('http://127.0.0.1:8888',timeout=2)
    print s.test()