使用 Paho Mqtt 客户端发布和订阅答案

【问题标题】：Publish and Subscribe with Paho Mqtt Client使用 Paho Mqtt 客户端发布和订阅
【发布时间】：2015-08-02 18:56:01
【问题描述】：

在我的树莓派上安装 paho mqtt 客户端并将我的 android 应用程序和 arduino 与我的代理连接后。我希望在收到来自应用程序或 arduino 客户端的消息后将此消息重新发布到此客户端，例如，如果我在收到“OFF”后收到“ON”，下一次发布将是“ON”“OFF”“ON”“OFF” “......或者我需要只是“开”或“关”

import paho.mqtt.client as mqtt
message = 'ON'
def on_connect(mosq, obj, rc):
    print("rc: " + str(rc))

def on_message(mosq, obj, msg):
    global message
    print(msg.topic + " " + str(msg.qos) + " " + str(msg.payload))
    message = msg.payload

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

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

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

mqttc = mqtt.Client()
# Assign event callbacks
mqttc.on_message = on_message
mqttc.on_connect = on_connect
mqttc.on_publish = on_publish
mqttc.on_subscribe = on_subscribe
# Connect
mqttc.connect("localhost", 1883,60)

# Start subscribe, with QoS level 0
mqttc.subscribe("f", 0)

# Publish a message
#mqttc.publish("hello/world", "my message")

# Continue the network loop, exit when an error occurs
rc = 0
while rc == 0:
   rc = mqttc.loop()
   mqttc.publish("f",message)
print("rc: " + str(rc))

【问题讨论】：

能否请您多解释一下您想要达到的目标，目前还不是很清楚。
请确保订阅 on_connect 回调，否则是一个错误。请参阅此处的入门部分：eclipse.org/paho/clients/python
我之前试过这个，和我一样，我的问题是发布方法不是订阅
@hardillb 我想重新发布从我订阅的主题收到的每条消息

标签： python raspberry-pi mqtt paho

【解决方案1】：

你可以研究这段代码并自己使用，

import RPi.GPIO as GPIO
import time
import paho.mqtt.client as mqtt

GPIO.setmode(GPIO.BOARD)

GPIO.setup(13,GPIO.IN)

mqttc=mqtt.Client()
mqttc.connect("iot.eclipse.org",1883,60)
mqttc.loop_start()

def reading1():
    a=GPIO.input(13)
    print(a)
    return a

while 1:
    t=reading1()
    (result,mid)=mqttc.publish("paho/temp_ab",t,2)
    time.sleep(1)

mqttc.loop_stop()
mqttc.disconnect()

【讨论】：

【解决方案2】：

好的，

以下代码将订阅主题 f 并重新发布主题 f2

import paho.mqtt.client as mqtt
message = 'ON'
def on_connect(mosq, obj, rc):
    mqttc.subscribe("f", 0)
    print("rc: " + str(rc))

def on_message(mosq, obj, msg):
    global message
    print(msg.topic + " " + str(msg.qos) + " " + str(msg.payload))
    message = msg.payload
    mqttc.publish("f2",msg.payload);

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

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

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

mqttc = mqtt.Client()
# Assign event callbacks
mqttc.on_message = on_message
mqttc.on_connect = on_connect
mqttc.on_publish = on_publish
mqttc.on_subscribe = on_subscribe
# Connect
mqttc.connect("localhost", 1883,60)


# Continue the network loop
mqttc.loop_forever()

【讨论】：

对不起兄弟，但我希望它具有相同的主题，因为我的目标是将收到的消息分享给订阅 f 而不是 f2 的客户（android，arduino）
在同一主题上重新发布相同的值是没有意义的，因为您最终会陷入持续的反馈循环。没有什么可以阻止 android 和 arduino 订阅同一个主题。
所以你认为我只需要订阅主题并在我的经纪人和我的所有客户中发布到另一个
您可能应该将命令与不同主题的状态分开

【解决方案3】：

我认为这可能会对您有所帮助。

import threading
import paho.mqtt.client as mqtt

def publish_1(client,topic):
    message="on"
    print("publish data")
    client.publish(topic,message)
    publish_1(client,topic)


broker="test.mosquitto.org"
topic_pub='/temperature123'
topic_sub='$SYS/#'

def on_connect(client, userdata, rc):
    print("Connected with result code "+str(rc))
    client.subscribe(topic_sub)


def on_message(client, userdata, msg):
    print(msg.topic+" "+str(msg.payload))

client = mqtt.Client()
client.on_connect = on_connect
client.on_message = on_message

client.connect(broker, 1883, 60)
thread1=threading.Thread(target=publish_1,args=(client,topic_pub))
thread1.start()

client.loop_forever()

此程序使用线程并传递客户端对象。但是这个递归发布的问题。

【讨论】：