ZeroMQ ROUTER套接字无法向REP套接字发送消息

Question

我使用ZeroMQ实现了一个简单的请求 - 回复架构和路由器。这适用于 PyZMQ版本2.1.11 。不幸的是，当我在 PyZMQ版本14.0.0 上测试它时，发送方（REQ）可以发送到路由器然后路由器收到它的消息并发送到接收方（REP）但接收方没有收到消息！ 当我将PyZMQ从2.1.11升级到14.0.0时，我遇到了这个问题。

REQ＆lt; - ＆gt; ROUTER＆lt; - ＆gt; REP

这是我的代码：

sender.py

import zmq
import time

if __name__=='__main__':
    context = zmq.Context()
    socket = context.socket(zmq.REQ)
    socket.setsockopt(zmq.IDENTITY, "S")
    socket.connect("tcp://127.0.0.1:6660")
    i = 0
    while True:
        i += 1
        socket.send("R", zmq.SNDMORE)
        socket.send("", zmq.SNDMORE)
        socket.send("Message: %d" % i)
        print("Message : %d sent" % i)
        fromAddr = socket.recv()
        empty = socket.recv()
        resp = socket.recv()
        print("%s received!" % str(resp))
        time.sleep(1)

router.py

import zmq
import time

if __name__=='__main__':
    context = zmq.Context()
    frontend = context.socket(zmq.ROUTER)
    frontend.bind("tcp://*:6660")

    poll = zmq.Poller()
    poll.register(frontend, zmq.POLLIN)

    while True:
        sockets = dict(poll.poll(100))
        if frontend in sockets:
            if sockets[frontend] == zmq.POLLIN:
                fromAddr = frontend.recv()
                empty = frontend.recv()
                toAddr = frontend.recv()
                empty = frontend.recv()
                msg = frontend.recv()
                print("Message received from %s must be send to %s [%s]" % (str$
                frontend.send(toAddr, zmq.SNDMORE)
                frontend.send("", zmq.SNDMORE)
                frontend.send(fromAddr, zmq.SNDMORE)
                frontend.send("", zmq.SNDMORE)
                frontend.send(msg)
                print("Message has been send to %s!" % str(toAddr))

receiver.py

import zmq
import time

if __name__=='__main__':
    context = zmq.Context()
    socket = context.socket(zmq.REP)
    socket.setsockopt(zmq.IDENTITY, "R")
    socket.connect("tcp://127.0.0.1:6660")
    while True:
        print("Wating for request...")
        toAddr = socket.recv()
        empty = socket.recv()
        req = socket.recv()
        print("%s received!" % str(req))
        socket.send(toAddr, zmq.SNDMORE)
        socket.send(empty, zmq.SNDMORE)
        socket.send("Reply to %s" % str(req))

当我使用这种架构时：

DEALER不会路由到多个接收者。 DEALER仅使用循环方法向接收方发送消息。如果可以使用ROUTER代替DEALER，则可以将消息路由到特定的接收者，并在这些消息之间进行循环。

Answer 1

ROUTER to REP socket是一个无效的组合，如下所述：http://zguide.zeromq.org/page:all#Request-Reply-Combinations

Answer 2

根据@nos所说的，ROUTER到REP是无效组合，但ROUTER到ROUTER套接字是有效的。简单地说，我将REP套接字更改为ROUTER！修改后的代码在这里：

import zmq

if __name__=='__main__':
    context = zmq.Context()
    socket = context.socket(zmq.ROUTER)       # Changed
    socket.setsockopt(zmq.IDENTITY, "R1")
    socket.connect("tcp://127.0.0.1:6660")
    while True:
        print("Wating for request...")

        me = socket.recv()       # New
        empty = socket.recv()    # New
        toAddr = socket.recv()
        empty = socket.recv()
        req = socket.recv()
        print("%s received!" % str(req))

        socket.send(me, zmq.SNDMORE)       # New
        socket.send(empty, zmq.SNDMORE)    # New
        socket.send(toAddr, zmq.SNDMORE)
        socket.send(empty, zmq.SNDMORE)
        socket.send("Reply to %s" % str(req))