[BUG] ObserverModelInstanceMixin not observable inside Celery task

[Alexsandr]

Describe the bug
Hello, my friend
Thank you for your hard work in creating this wonderful library.
I have one small question
The behavior I'm seeing is that when I run the celery task, Observer doesn't work as I expect. When changing a task through the Django admin panel, everything works fine, the model change is immediately sent to my websocket, but when I run the task in celery, the notifications disappear.
I tried to find the reason for 2 days but without success
I hope you can help me, my friend!

To Reproduce
I won’t go into too much depth, I’ll describe the main points, they will be enough to understand the essence of the problem

docker compose
`
...etc...
redis:
image: redis:alpine
container_name: redis_development
ports:
- "6379:6379"
celery:
build:
context: .
dockerfile: docker/django/development/Dockerfile
container_name: worker_development
volumes:
- .:/fast-check-up
command: celery -A config.celery worker --loglevel=INFO
env_file:
- docker/django/development/.env.dev
depends_on:
- redis
- django
restart: unless-stopped

...etc...
`

Settings
`
CELERY_BROKER_URL = 'redis://redis:6379/0'
CELERY_RESULT_BACKEND = 'redis://redis:6379/0'

Channels

CHANNEL_LAYERS = {
'default': {
'BACKEND': 'channels_redis.core.RedisChannelLayer',
'CONFIG': {
'hosts': ['redis://redis:6379/0'],
},
},
}
`

Consumer
`class StudyConsumer(ObserverModelInstanceMixin, GenericAsyncAPIConsumer, PatchModelMixin):
queryset = Study.objects.all()
serializer_class = StudiesSerializer

@model_observer(Study)
async def model_change(self, message, **kwargs):
    await self.send_json(message)

@action()
def retrieve(self, **kwargs) -> Tuple[ReturnDict, int]:
    instance = self.get_object(**kwargs)
    serializer = self.get_serializer(instance=instance, action_kwargs=kwargs)
    return serializer.data, status.HTTP_200_OK

@action()
def patch(self, data: dict, **kwargs) -> Tuple[ReturnDict, int]:
    instance = self.get_object(data=data, **kwargs)
    serializer = self.get_serializer(instance=instance, data=data, action_kwargs=kwargs, partial=True)
    serializer.is_valid(raise_exception=True)
    self.perform_patch(serializer, **kwargs)
    if getattr(instance, "_prefetched_objects_cache", None):
        instance._prefetched_objects_cache = {}

    if data['status'] == Study.PROCESSING:
        run.delay(instance.pk) # Here i call the task

    return serializer.data, status.HTTP_200_OK`

Celery task:
def run(pk: str): study_record = Study.objects.get(id=pk) study_record.completed_at = datetime.datetime.now() study_record.status = Study.COMPLETED study_record.save(update_fields=['processing_time', 'result', 'status', 'started_at', 'completed_at']) return 'A processing celery task was complete!'

Expected behavior
I'm guessing that changes inside the celery task should also fire an observer trigger and be sent to Websocket

• OS: Windows 10. Projects starts in Docker with docker compose

Thanks for any answer
Good luck!

[Alexsandr]

class TestAppConfig(AppConfig):
    default_auto_field = 'django.db.models.BigAutoField'
    name = 'testapp'

    def ready(self):
        from OtherApp import consumers