Register class-based task in

Question

I am using Celery version 4.0.2.

Compared to previous versions of Celery, it seems that class-based tasks are not registered automatically (i.e. if you configured auto-discovery).

However, I do not even achieve registering a class-based task manually.

According to the Celery change log:

http://docs.celeryproject.org/en/latest/changelog.html#version-4-0-1

since version 4.0.1 it should be possible to register the task manually:

from celery import Celery, Task
app = Celery()

class CustomTask(Task):

    def run(self):
        return 'hello'

app.register_task(CustomTask())

But this does not seem to work. Does anyone know how to achieve this?

I tried a few suggestions which are being discussed (apart from integrating a custom task loader mentioned in https://github.com/celery/celery/issues/3744):

Register Celery Class-based Task

https://github.com/celery/celery/issues/3615

https://github.com/celery/celery/issues/3744

score 2 · Accepted Answer · answered Jul 06 '17 at 14:46

2

Almost there! You need to call delay() on the task that you registered.

This would work:

from celery import Celery, Task

app = Celery()


class CustomTask(Task):
    def run(self):
        return 'hello'


task = CustomTask()
app.register_task(task)

task.delay()

answered Jul 06 '17 at 14:46

Jurrian

532
4
14

But it's not worked when importing `task` on other files. – harukaeru Aug 31 '17 at 09:39

score 0 · Answer 2 · answered Jul 09 '17 at 16:24

If You need shared_task decorator:

from celery import Task, shared_task

class CustomTask(Task):
    def process(self):
        return 'hello'

@shared_task(bind=True, base=CustomTask)
def custom(self):
    self.process()

process is a custom name that starts task (decorator overrides run method)

bind=True binds function to a class instance

base=CustomTask sets a base class for a task

Register class-based task in

2 Answers2