Can the callbacks generated by subscribing to notifications last forever

[Luo-XiuFeng]

while(True): await self.client.start_notify(self.noticeuuid, notification_handler())

1. When I find the uuid of the notification and enable notification, I can get a notification_handler to process, but in order to keep getting the message, I have to use while(true) to always enable notification. Is there a better way?
2. There is a serious time problem with while(true). My Bluetooth is sent once a second, but when I print the time in the notification_handler callback, I find that notice is not called strictly in 1 second? How to solve it?

[dlech]

What event should trigger the program to end? For example, maybe the program should end when the device disconnects. The you can use asyncio.Event to wait for that event. See examples/disconnect_callback.py.

[dlech]

start_notify() should only be called once when the device is connected. Notifications will stay enabled until stop_notify() is called or the device disconnects.

[Luo-XiuFeng]

start_notify() should only be called once when the device is connected. Notifications will stay enabled until stop_notify() is called or the device disconnects.

I want to always subscribe to notifications and be able to always receive callbacks to process my data, but after subscribing only through await client.start_notify(CHARACTERISTIC_UUID, notification_handler), I can only receive notifications once and never receive them again. I didn’t call start_stop, nor did I call it unless the device was disconnected。
Suppose I did not comment out the while (true), I will always receive the callback, but the number of callbacks does not match the sending frequency of the Bluetooth data, the callback is very slow, when I comment out the while (true), I will only receive it once Or a second callback.

    async def run(self,debug=0):
        
        log = logging.getLogger(__name__)
        if debug:   
            import sys

        async with BleakClient(self.bleMac) as self.client:
            x = await self.client.is_connected()
            keyuu = '6e'
            for service in self.client.services:
                print(service.uuid)
                if keyuu not in service.uuid:
                    continue

                for char in service.characteristics:
                    print(char.properties)
                    if 'notify' not in char.properties:
                        continue
                    else:
                        print(char.uuid)
                        self.noticeuuid = char.uuid
                        try:
                            value = bytes(await self.client.read_gatt_char(char.uuid))
                        except Exception as e:
                            value = str(e).encode()
                    

                    for descriptor in char.descriptors:
                        value = await self.client.read_gatt_descriptor(descriptor.handle)
                        # await asyncio.sleep(1)
                        #await self.client.stop_notify(CHARACTERISTIC_UUID)

            # while(True):
                await self.client.start_notify(self.noticeuuid,self.notification_handler)

[Luo-XiuFeng]

start_notify() should only be called once when the device is connected. Notifications will stay enabled until stop_notify() is called or the device disconnects.

Hello, what should I do to keep being notified at all times? Or is there an example for reference?