subs.py

# Copyright (c) 2010-2013 Roger Light <roger@atchoo.org>
#
# All rights reserved. This program and the accompanying materials
# are made available under the terms of the Eclipse Distribution License v1.0
# which accompanies this distribution.
#
# The Eclipse Distribution License is available at
#   http://www.eclipse.org/org/documents/edl-v10.php.
#
# Contributors:
#    Roger Light - initial implementation
# Copyright (c) 2010,2011 Roger Light <roger@atchoo.org>
# All rights reserved.

# This shows a simple example of an MQTT subscriber.

import context  # Ensures paho is in PYTHONPATH
import paho.mqtt.client as mqtt
import time
import re
import paho.mqtt.publish as publish

def on_connect(mqttc, obj, flags, rc):
    print("rc: " + str(rc))


def on_message(mqttc, obj, msg):
    sendtime = re.findall(r"\d{13}", str(msg.payload))
    payload=msg.payload
    publish.single("response", payload, hostname="172.18.163.31", retain=False)#change your topic and server address
    print("send " + str(sendtime) + " out")

def on_publish(mqttc, obj, mid): 
    print("mid: " + str(mid))


def on_subscribe(mqttc, obj, mid, granted_qos):
    print("Subscribed: " + str(mid) + " " + str(granted_qos))


def on_log(mqttc, obj, level, string):
    print(string)


# If you want to use a specific client id, use
mqttc = mqtt.Client("client2")
# but note that the client id must be unique on the broker. Leaving the client
# id parameter empty will generate a random id for you.
mqttc.on_message = on_message
mqttc.on_connect = on_connect
mqttc.on_publish = on_publish
mqttc.on_subscribe = on_subscribe
# Uncomment to enable debug messages
mqttc.on_log = on_log
mqttc.connect("172.18.163.31", 1883, 60)
mqttc.subscribe("request", 0)

mqttc.loop_forever()