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I am trying to publish a message(it's like broadcast when using raw sockets) to my subnet with a known port but at subscriber end, the message is not received. The idea is the IP address of the first machine should not be known to the second machine that's why I am using broadcast IP. With UDP or TCP raw socket, it works but I am trying to learn pub-sub pattern not sure how to incorporate that idea.

This is my codes:

Publisher:

import zmq
import sys
import time
context=zmq.Context()
socket=context.socket(zmq.PUB)
socket.bind("tcp://192.168.1.255:5677")
while True:
    data='hello'.encode()
    socket.send(data)
    #time.sleep(1)

Subscriber:

context=zmq.Context()
    sub=context.socket(zmq.PUB)
    sub.setsocketopt(zmq.SUBSCRIBE, "".encode())
    sub.connect('tcp://192.168.1.255:5677')
    sub.recv()
    print(sub.recv())

In terms of raw UDP, I wrote a code which works perfectly.

broadcast:

def broadcast(Host,port):
    #send bd
    sock=socket.socket(socket.AF_INET,socket.SOCK_DGRAM)
    msg=get_ip_data("wlp3s0")
    sock.setsockopt(socket.SOL_SOCKET, socket.SO_BROADCAST, 1)
    time.sleep(1.5)
    # print("yes sending", client)

    sock.sendto(msg.encode(), (Host,port))

recv:

def broadcast_recv():
    #listen bd
    sock=socket.socket(socket.AF_INET,socket.SOCK_DGRAM)
    sock.bind((get_bd_address("wlp1s0"),12345))
    # receive broadcast
    msg, client = sock.recvfrom(1024)
    a=(msg.decode())
    print(a)
Benyamin Jafari
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KSp
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2 Answers2

3

It seems you forgot the zmq.SUB in the subscriber side. Also you used sub.setsocketopt() instead of sub.setsockopt().

Try it:

Publisher:

import zmq
import time

context = zmq.Context()
socket = context.socket(zmq.PUB)
socket.bind("tcp://*:5677")  # Note.

while True:
    socket.send_string('hello')
    time.sleep(1)

Subscriber:

context = zmq.Context()
sub=context.socket(zmq.SUB)  # Note.
sub.setsockopt(zmq.SUBSCRIBE, b"")  # Note.
sub.connect('tcp://192.168.1.255:5677')

while True:
    print(sub.recv())

[NOTE]:

  • You can also change the .bind() and .connect() in subscriber and publisher with your policy. (This post is relevant).
  • Make sure that 5677 is open in the firewall.
  • socket.bind("tcp://*:5677") or socket.bind("tcp://0.0.0.0:5677") is broadcasting trick.
Benyamin Jafari
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  • Hi, thanks for your script. I was also thinking if broadcasting is possible in zmq. @benyamin jafari – KSp Nov 23 '18 at 16:53
  • @keerthana `socket.bind("tcp://*:5677")` or `socket.bind("tcp://0.0.0.0:5677")` is broadcasting trick. – Benyamin Jafari Nov 23 '18 at 16:55
  • Thanks, you answered my pub-sub doubts so checked it as answered, though i want to broadcast to subnet address. So i cannot use 0.0.0.0 address but use 192.X.X.255 address and where the subscriber does not receive the data. :( – KSp Nov 24 '18 at 10:16
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I think the problem is that the SUB socket cannot register itself with the PUB socket. Even though in-concept the data only goes from PUB to SUB, in reality, there are also control messages (e.g. subscription topics), being sent back to the PUB.

If your netmask is 255.255.255.0, this will probably not work as expected.

user3097732
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