serve_forever在对面收通知发请求前开始工作就行
看代码似乎不用太担心：

    def serve_forever(self, poll_interval=0.5):
        """Handle one request at a time until shutdown.

        Polls for shutdown every poll_interval seconds. Ignores
        self.timeout. If you need to do periodic tasks, do them in
        another thread.
        """
        self.__is_shut_down.clear()
        try:
            # XXX: Consider using another file descriptor or connecting to the
            # socket to wake this up instead of polling. Polling reduces our
            # responsiveness to a shutdown request and wastes cpu at all other
            # times.
            with _ServerSelector() as selector:
                selector.register(self, selectors.EVENT_READ)

                while not self.__shutdown_request:
                    ready = selector.select(poll_interval)
                    # bpo-35017: shutdown() called during select(), exit immediately.
                    if self.__shutdown_request:
                        break
                    if ready:
                        self._handle_request_noblock()

                    self.service_actions()
        finally:
            self.__shutdown_request = False
            self.__is_shut_down.set()


【 在 JulyClyde 的大作中提到: 】
: 标  题: Re: 如何自己开一个http服务器并通知别人来访问？
: 发信站: 水木社区 (Fri Jun 21 11:54:00 2024), 转信
:
:
: 【 在 vwx 的大作中提到: 】
: : 真走火入魔
: : 搜到的代码
: : import http.server
: : import socketserver
: : PORT = 8080
: : Handler = http.server.SimpleHTTPRequestHandler
: : with socketserver.TCPServer(("", PORT), Handler) as httpd:
: :     ### 写通知
: :     httpd.serve_forever()
: 这不就是我说的那种明显的错误么？
:
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:
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