python - How to handle multiple publishers on same port in zmq? -


this question has been asked before, here. have exact same problem. want publish bunch of different processes, , use same port every time.

i tried solution presented in answer, did not work me. error

    file "/usr/local/lib/python2.7/multiprocessing/process.py", line 258, in _bootstrap     self.run()   file "/usr/local/lib/python2.7/multiprocessing/process.py", line 114, in run     self._target(*self._args, **self._kwargs)   file "/home/akay/afk/multi.py", line 18, in to_zmq     socket.connect("tcp://*:%s" % port)   file "zmq/backend/cython/socket.pyx", line 478, in zmq.backend.cython.socket.socket.connect (zmq/backend/cython/socket.c:4308) zmqerror: invalid argument 

my code this, taken straight example in zmq docs here , here:

# socket talk server port = '5556' context = zmq.context() socket = context.socket(zmq.sub) print "listening stream...", m socket.bind("tcp://localhost:%s" % port) #change connect bind, per answer above socket.setsockopt(zmq.subscribe, topicfilter) 

i using python 2.7, , recent version of zmq. idea might doing wrong?

well, error clear:

    [...]     socket.connect("tcp://*:%s" % port)     [...] zmqerror: invalid argument 

you can't connect *, must specify ip address (the server ip address). if both client , server run on single machine, try localhost or 127.0.0.1.


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