Question

我正在尝试在Django上实现Paho MQTT，但Django在引发任何异常后停止接收/发送MQTT消息。

我正在使用Paho MQTT客户端1.3.0以及Django 1.10.8，Python 3.6.2

以下是我的MQTT设置：

mqtt.py

from django.conf import settings

import paho.mqtt.client as mqtt

SUB_TOPICS = ("device/vlt", "device/auth", "device/cfg", "device/hlt", "device/etracker", "device/pi")
RECONNECT_DELAY_SECS = 2


# The callback for when the client receives a CONNACK response from the server.
def on_connect(client, userdata, flags, rc):
    print("Connected with result code " + str(rc))

    # Subscribing in on_connect() means that if we lose the connection and
    # reconnect then subscriptions will be renewed.
    for topic in SUB_TOPICS:
        client.subscribe(topic, qos=0)


# The callback for when a PUBLISH message is received from the server.
def on_message(client, userdata, msg):
    print(msg.topic + " " + str(msg.qos) + " " + str(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)


def on_disconnect(client, userdata, rc):
    client.loop_stop(force=False)
    if rc != 0:
        print("Unexpected disconnection: rc:" + str(rc))
    else:
        print("Disconnected: rc:" + str(rc))


client = mqtt.Client()
client.on_connect = on_connect
client.on_message = on_message
client.on_publish = on_publish
client.on_subscribe = on_subscribe
client.on_disconnect = on_disconnect

client.username_pw_set(username, password)
client.connect(<settings>)

apps.py

class CoreConfig(AppConfig):
    name = 'untitled.core'
    verbose_name = "Core"

    def ready(self):
        from . import mqtt
        mqtt.client.loop_start()

代码灵感：Paho MQTT client connection reliability (reconnect on disconnection)

Answer 1

我的团队也遇到了这个问题，我们通过创建一个继承自CustomMqttClient的{{1}}并覆盖mqtt.Client函数来解决此问题。

_handle_on_message(self, message)

然后我们执行此操作而不是使用class CustomMqttClient(mqtt.Client): def _handle_on_message(self, message): try: super(ChatqMqttClient, self)._handle_on_message(message) except Exception as e: error = {"exception": str(e.__class__.__name__), "message": str(e)} self.publish("device/exception", json.dumps(error))：

mqtt.Client()

这会捕获所有异常，并实际将其发布到我们的主题client = CustomMqttClient() client.on_connect = on_connect client.on_message = on_message。我们的其他服务实际上可以订阅它，并从中获得一些有用的信息。

Django在引发任何异常后停止接收/发送MQTT消息

1 个答案: