How do I know if receive topic comes from the same subscribe?

[jeffreykira]

I subscribe a topic with plus symbol (LIGHT/+/TURNON)
So I will receive these topics:

LIGHT/ID_1/TURNON
LIGHT/ID_2/TURNON

I hope to receive these topic can do the same thing.
How do I know that LIGHT/ID_1/TURNON and LIGHT/ID_2/TURNON are from the same subscribe?

following is my sample code:

TOPIC_LIST = []

def turn_on_tv(client, topic, message, qos, properties):
    # do something

def turn_on_light(client, topic, message, qos, properties):
    # do something

def init_subscribe(client):
    global TOPIC_LIST
    TOPIC_LIST = [
        {'topic': 'TV/TURNON'., 'qos': 1, 'method': turn_on_tv},
        {'topic': 'LIGHT/+/TURNON', 'qos': 1, 'method': turn_on_light},
    ]

    for t in TOPIC_LIST:
        client.subscribe(t['topic'], t['qos'])

def on_message(client, topic, payload, qos, properties):
    for t in TOPIC_LIST:
        if topic == t['topic']:    # How do I know?
            t['method'](client, topic, message, qos, properties)
            break

[Lenka42]

@jeffreykira For now there are two possibilities:

1. you can match received message topic with regexp, like this:

pattern = r'LIGHT/(?P<device_id>\d+)/TURNON'
match = re.fullmatch(pattern, topic)
if match is not None:
    device_id = match.group('device_id')
    # do the turn on logic here

2. or if you use broker supporting MQTT 5.0 version, you can use subscription identifiers for this.

    TOPIC_LIST = [
        {'topic': 'TV/TURNON'., 'qos': 1, 'method': turn_on_tv, 'subscription_identifier': 1},
        {'topic': 'LIGHT/+/TURNON', 'qos': 1, 'method': turn_on_light, 'subscription_identifier': 2},
    ]

    for t in TOPIC_LIST:
        # pass it as kwarg to subscribe method
        client.subscribe(t['topic'], t['qos'], subscription_identifier=t['subscription_identifier'])

on_message handler will receive messgae with property subscription_identifier, which is list:

def on_message(client, topic, payload, qos, properties):
    if 2 in properties['subscription_identifier']:
        # do the turnon logic here

[jeffreykira]

@Lenka42 Thank you very much for your reply, I will use the second one.