Python服务器在闲置一段时间后自行关闭

时间:2012-10-08 13:16:51

标签: python timeout

最简单的创建Python服务器(XML-RPC Server)的方法是什么?它会在空闲后关闭自己?

我想这样做,但我不知道该怎么做todo:

from SimpleXMLRPCServer import SimpleXMLRPCServer

# my_paths variable
my_paths = []

# Create server
server = SimpleXMLRPCServer(("localhost", 7789))
server.register_introspection_functions()

# TODO: set the timeout
# TODO: set a function to be run on timeout

class MyFunctions:
    def add_path(paths):
        for path in paths:
            my_paths.append(path)
        return "done"

    def _dispatch(self, method, params):
        if method == 'add_path':
            # TODO: reset timeout
            return add_path(*params)
        else:
            raise 'bad method'

server.register_instance(MyFunctions())

# Run the server's main loop
server.serve_forever()

我还尝试在示例here之后探索signal.alarm(),但它不会在Windows下向我投掷AttributeError: 'module' object has no attribute 'SIGALRM'

感谢。

1 个答案:

答案 0 :(得分:1)

您可以创建自己的扩展SimpleXMLRPCServer的服务器类,以便在空闲时关闭。

class MyXMLRPCServer(SimpleXMLRPCServer):
    def __init__(self, addr):
        self.idle_timeout = 5.0 # In seconds
        self.idle_timer = Timer(self.idle_timeout, self.shutdown)
        self.idle_timer.start()
        SimpleXMLRPCServer.__init__(self, addr)

    def process_request(self, request, client_address):
        # Cancel the previous timer and create a new timer
        self.idle_timer.cancel()
        self.idle_timer = Timer(self.idle_timeout, self.shutdown)
        self.idle_timer.start()
        SimpleXMLRPCServer.process_request(self, request, client_address)

现在,您可以使用此类来创建服务器对象。

# Create server
server = MyXMLRPCServer(("localhost", 7789))
server.register_introspection_functions()