2017-01-14 499 views
0

我正在使用MQTT來監視我訂閱的一些頻道。現在我想實現發送消息作爲對狀態的反應。我用下面的代碼運行它,在那裏我只是在on_message回調中做出反應(代碼1在最後)。但是這段代碼使用MQTT:如何發送消息而無需調用loop_forever()

loop_forever() 

在阻塞的主要代碼中。

我想要做的是隻發送一條消息到MQTT。

import paho.mqtt.client as mqtt 

if __name__ == "__main__": 

    mqtt_client = mqtt.Client() 
    mqtt_client.connect("192.168.178.204", 1883, 60) 
    mqtt_client.username_pw_set(username="test", password="test") 

    mqtt_client.publish(topic='TEST', payload='CCCCCCCCC', retain=False) 
    mqtt_client.loop_write() 
    # mqtt_client.loop() 
    # mqtt_client.loop_start()  

    mqtt_client.disconnect() 

我如何將消息發送給MQTT不擋住過程:當我嘗試以下方法(與所有不同的循環功能),沒有什麼是由MQTT服務器接收?


代碼1:

import paho.mqtt.client as mqtt 

def on_connect(client, userdata, rc): 
    topic_list = [("TEST_MS", 1)] 
    if rc == 0: 
     print("Successful connected and subscribed to: {}".format(topic_list)) 
    client.subscribe(topic_list) 


def on_message(client, userdata, msg): 
    print(msg.payload) 
    client.publish(topic='TEST_MS2', payload=msg.payload, retain=False) 


def on_publish(client, userdata, mid): 
    print("message published") 


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

if __name__ == "__main__": 

    mqtt_client = mqtt.Client() 

    mqtt_client.on_connect = on_connect 
    mqtt_client.on_message = on_message 
    mqtt_client.on_publish = on_publish 
    mqtt_client.on_subscribe = on_subscribe 

    mqtt_client.connect("192.168.178.204", 1883, 60) 
    mqtt_client.username_pw_set(username="test", password="test") 

    # mqtt_client.publish(topic='TEST_MS', payload='CCCCCCCCC', retain=False) 
    mqtt_client.loop_forever() 

    mqtt_client.disconnect() 

回答

1

如果你想發送只是一個單一的消息,並退出使用API​​專爲。 Docs here

import paho.mqtt.publish as publish 

publish.single("paho/test/single", "payload", hostname="iot.eclipse.org") 
+0

非常好,非常感謝你 – mstuebner

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