MQTT Publisher in Python

Question

I am publishing some data from a bme280 sensor to a mosquitto broker. I want do this every 3 seconds.

The readings from sensor are always ok but the publish to mosquitto very often loose some message. I can check this with a external MQTT subscrive client.

Where is the error? Here my code,It was assembled from some examples so sorry for non very good code.... Thanks.

#!/usr/bin/python3

import smbus2
import bme280
from paho.mqtt import client as mqttClient
import random
import time

lst2 = [None] * 2
lst2[0] = 'Temperatura stanzino'
lst2[1] = 'Umidita stanzino'
parolatot2 = [None] * 2

def on_connect(client, userdata, flags, rc):
        if rc == 0:
                pass
                global Connected
                Connected = True
        else:
                pass

Connected = False
broker_address= "192.168.1.250"
port = 1883
user = "xxx"
password = "xxx"
topic = "python/mqtt/"
client = mqttClient.Client("Python")
client.username_pw_set(user, password=password) 
client.on_connect= on_connect 
client.connect(broker_address, port=port)      
client.loop_start() 

while Connected != True:
        time.sleep(0.1) 

def calcoloTotale():
        msg2_count = 0
        port = 1
        address = 0x76
        bus = smbus2.SMBus(port)
        calibration_params = bme280.load_calibration_params(bus, address)
        data = bme280.sample(bus, address, calibration_params)

        parolatot2[0] = round(data.temperature,1)
        parolatot2[1] = round(data.humidity,1)

        print (parolatot2)
        while msg2_count < 2:
                topicnew2 = topic + lst2[msg2_count]
                time.sleep(.001)
                msg2 = f"{parolatot2[msg2_count]}"
                client.publish(topicnew2, msg2)
                msg2_count += 1

try:
        while True:
                calcoloTotale()
                time.sleep(3.0)
except:
        print ("errore")
        client.disconnect()
        client.loop_stop()

Answer 1

The client.publish method never throws an error. If you want to make sure that the message has really been published (and/or understand why not), I would recommend the following:

message_info = client.publish(topic, message, qos=1)
message_info.wait_for_publish(timeout=2) # will throw error if message isn't published within timeout.

Notes:

• qos=1 : use at least qos 1 if you want to make sure your message got published at least once.
• Make sure to use the paho-mqtt >= 1.6.1 to be able to use the wait_for_publish method.

Answer 2

Thanks my friend,i have changed approach to this.I'm using now "publish.single" istruction.So my code was reduced a lot. Now just a single shot to public and no more loose message. I have used this:

import paho.mqtt.publish as publish

publish.single(topicnew2, msg2, qos=1, retain=False, hostname="192.168.1.250", auth={'username':" ", 'password':" "})

But now another issue,the qos is always 0 in the client.Some idea? Thanks