Question

我使用xmlrpc服务器运行简单示例，然后在键盘上按Ctrl-C :)。

from SimpleXMLRPCServer import SimpleXMLRPCServer
from time import sleep
import threading,time

class 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 + 1

ts = Test()
ts.start()
server =  SimpleXMLRPCServer(("localhost",8888))
server.register_instance(ts)
server.serve_forever()

按键盘后出现

错误：

  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 ServerProxy
r=ServerProxy("http://localhost:8888")
print r.test()

等待连接没有错误或警告。在这种情况下如何打破连接？也许这个例子不正确？

Answer 1

使用超时：

Set timeout for xmlrpclib.ServerProxy

修改

此处链接的答案与Python 2.7不兼容。这是修改后的代码（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()

Answer 2

使你的Test实例守护进程，在主线程退出时退出：

ts = Test()
ts.setDaemon(True)
ts.start()

问题是为什么需要将线程注册为XML-RPC处理程序。

xmlrpc服务器的问题

2 个答案: