使用Paho MQTT在Python中订阅MQTT时发生线程问题

2023年2月5日 239次阅读

我有一个
python程序,它监听MQTT主题并需要处理消息.我从命令行指定了许多参数来不同地评估消息.

import argparse
import datetime
import json

import paho.mqtt.client as mqtt

### MQTT Functions
def on_connect(mqttc, obj, flags, rc):
    print("Connected! - " + str(rc))

def on_message(mqttc, obj, msg):
    print(msg.topic+" "+str(msg.qos)+" "+str(msg.payload))

def on_publish(mqttc, obj, mid):
    print("Published! "+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 __name__ == "__main__":
    # Handle args
    parser = argparse.ArgumentParser(
        description='This is to be usedin conjunction with the WifiScanner on a Raspberry Pi')
    parser.add_argument('--topic', metavar='base/sub', type=str, nargs='?', help='Full topic to listen to. (Example "proximity/sensor")', default="proximity/#")
    parser.add_argument('--host', metavar='url', type=str, nargs='?',
                        help='UQL of MQTT server.')
    parser.add_argument('--graph', metavar='True/False', type=bool, nargs='?', help='Whether to print the data.', default=True)
    parser.add_argument('--timeout', metavar='sec', type=int, nargs='?', help='How long the device will be remembered', default=10)
    args = parser.parse_args()
    # MQTT
    mqttc = mqtt.Client()
    # 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(args.host, 1883, 60)
    mqttc.subscribe(args.topic, 0)
    # Start to listen    
    while True:
        print mqttc.loop()

这个问题是,我看不到将命令行参数传递给on_message回调的简单方法.所以我尝试使用.loop的返回值.但是,当我尝试使用Ctrl Z退出时(只有键盘中断有效),它不会退出MQTT线程并让它们继续运行.

文档和示例没有关于如何处理on_message回调之外的消息以及如何干净地退出的示例.

因此,非常感谢任何解决此问题的帮助.

提前致谢

最佳答案 您可以使用Client()构造函数中的userdata参数.它最终被传递给每个回调.

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