python - ZeroMQ Pub/Sub 最简单的例子不起作用——我做错了什么？

Question

在 example in zmq documentation 之后，这是我的看法，稍作改动。

发布者.py

import zmq
import random
import time

port = "5556"

context = zmq.Context()
socket = context.socket(zmq.PUB)
socket.bind("tcp://127.0.0.1:%s" % port)


while True:
    topic = random.randrange(9999,10005)
    messagedata = random.randrange(1,215) - 80
    print "%d %d" % (topic, messagedata)
    socket.send("%d %d" % (topic, messagedata))
    time.sleep(1)

订阅者.py

import sys
import zmq

port = "5556"

context = zmq.Context()
socket = context.socket(zmq.SUB)

print "Collecting updates from weather server..."
socket.connect ("tcp://127.0.0.1:%s" % port)


while True:
    print socket.recv()

然后我从终端执行它们，当发布者发布时(即打印出行)，订阅者永远不会退出 socket.recv()

Answer 1

添加

socket.setsockopt(zmq.SUBSCRIBE, "")

订阅者，成功了!

这是更新后的 subscriber.py

import sys
import zmq

port = "5556"

context = zmq.Context()
socket = context.socket(zmq.SUB)
socket.setsockopt(zmq.SUBSCRIBE, "")

print "Collecting updates from weather server..."
socket.connect ("tcp://127.0.0.1:%s" % port)

while True:
    print socket.recv()