MQTT問題:蟒蛇MQTT腳本來發送和接收消息
嗨,我試圖建立多樹莓派之間的MQTT網絡(從2開始)。 我有一個覆蓋有熱敏電阻傳感器的樹莓派(RPi-A)MQTT客戶端,以及一個作爲我的網絡集線器的覆盆子(RPi-B)MQTT經紀人/客戶端。 通過python腳本,我想每隔30分鐘從RPi-A通過MQTT發送溫度到主題傳感器/數據並由RPi-B接收。 當RPi-B通過主題傳感器/數據接收到來自RPi-A的消息時,我希望它通過MQTT主題傳感器/指令向RPi-A發送指令。 以下是我的腳本,到目前爲止RPi-A可以發送消息,RPi-B可以接收它們,但是我無法弄清楚RPi-B如何響應。
基本上,我想了解的是,MQTT設備可能同時充當代理和客戶端嗎? 而且,客戶端是否可以發送和接收消息,如果有的話如何通過python實現以上所有內容? 我讀過很多博客,官方MQTT文章和paho模塊文檔(對我來說這很難理解),但仍然無法弄清楚。你的幫助將是最有用的/讚賞的。
代碼RPI-A(與熱敏電阻傳感器):
from sense_hat import SenseHat
import time
import paho.mqtt.client as mqtt
import paho.mqtt.publish as publish
sense = SenseHat()
Broker = "192.168.1.252"
sub_topic = "sensor/instructions" # receive messages on this topic
pub_topic = "sensor/data" # send messages to this topic
############### sensehat inputs ##################
def read_temp():
t = sense.get_temperature()
t = round(t)
return t
def read_humidity():
h = sense.get_humidity()
h = round(h)
return h
def read_pressure():
p = sense.get_pressure()
p = round(p)
return p
def display_sensehat(message):
sense.show_message(message)
time.sleep(10)
############### MQTT section ##################
# when connecting to mqtt do this;
def on_connect(client, userdata, flags, rc):
print("Connected with result code "+str(rc))
client.subscribe(sub_topic)
# when receiving a mqtt message do this;
def on_message(client, userdata, msg):
message = str(msg.payload)
print(msg.topic+" "+message)
display_sensehat(message)
def publish_mqtt(sensor_data):
mqttc = mqtt.Client("python_pub")
mqttc.connect(Broker, 1883)
mqttc.publish(pub_topic, sensor_data)
#mqttc.loop(2) //timeout = 2s
def on_publish(mosq, obj, mid):
print("mid: " + str(mid))
client = mqtt.Client()
client.on_connect = on_connect
client.on_message = on_message
client.connect(Broker, 1883, 60)
while True:
sensor_data = [read_temp(), read_humidity(), read_pressure()]
publish.single("monto/solar/sensors", str(sensor_data), hostname = Broker)
time.sleep(1*60)
代碼RPI-B(網絡集線器):
import time
import paho.mqtt.client as mqtt
import paho.mqtt.publish as publish
Broker = "192.168.1.252"
sub_topic = "sensor/data" # receive messages on this topic
pub_topic = "sensor/instructions" # send messages to this topic
# mqtt section
# when connecting to mqtt do this;
def on_connect(client, userdata, flags, rc):
print("Connected with result code "+str(rc))
client.subscribe(sub_topic)
# when receiving a mqtt message do this;
def on_message(client, userdata, msg):
message = str(msg.payload)
print(msg.topic+" "+message)
publish_mqtt(‘got your message’)
# to send a message
def publish_mqtt(sensor_data):
mqttc = mqtt.Client("monto_hub")
mqttc.connect(Broker, 1883)
mqttc.publish(pub_topic, "this is the master speaking")
#mqttc.loop(2) //timeout = 2s
def on_publish(mosq, obj, mid):
print("mid: " + str(mid))
client = mqtt.Client()
client.on_connect = on_connect
client.on_message = on_message
client.connect(Broker, 1883, 60)
client.loop_forever()
是你看到什麼實際的錯誤?代碼看起來是正確的(RPI-A在循環之前不需要任何MQTT客戶端代碼,因爲您使用的是'publish.single'),並且RPI-B代碼乍看之下看起來不錯。 – hardillb
@hardillb感謝您的回覆,我希望RPi-A接收來自RPi-B的mqtt消息以及發送它們,所以我想我需要客戶端mqtt代碼。我本身沒有出現錯誤,但是RPi-B似乎沒有發送消息來響應RPi-A。 –
對不起,錯過了那一點。提供的答案 – hardillb