多个ESP8266订阅一个RPI 3

Question

我无法将 2 个（及更多）ESP8266 订阅给一个 RPi3 代理。我用过：

import paho.mqtt.client as mqtt
import datetime

mqtt_topics = ["esp8266-1", "esp8266-2"]

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

    for topic in mqtt_topics:
     client.subscribe(topic)

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

client = mqtt.Client()
client.on_connect = on_connect
client.connect('localhost', 1883, 60)

try:
    for topic in mqtt_topics:
        client.on_message = on_message
    client.loop_forever()
except KeyboardInterrupt:
        print("CTRL-C: Terminating program.")

它有效，但仅适用于 1 个 esp。如果我连接 esp8266-1 它会检查它的值，然后我连接 esp8266-2 以获得它的值并且 esp8266-1 不再可用（即使我打开 esp8266-2）。

如何同时订阅esp8266？我一次只能吃一个。

Answer 1

代码需要做一些改动：

import paho.mqtt.client as mqtt
import datetime

mqtt_topics = [("esp8266-1",0), ("esp8266-2",0)]

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

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

client = mqtt.Client()
client.on_connect = on_connect
client.on_message = on_message
client.connect('localhost', 1883, 60)

try:
  client.loop_forever()
except KeyboardInterrupt:
  print("CTRL-C: Terminating program.")

但是，如果一个客户端正常工作，而第二个强制第一个客户端断开连接，那么这听起来就像两个客户端的 clientID 相同。 clientIDs 需要对所有客户端都是唯一的。

多个ESP8266订阅一个RPI 3

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