CODEWITHSUNDEEP
pythoncelery
Yaroslav Melnichuk
Yaroslav Melnichuk

Reputation: 850

Do something on every Celery Task run

Assume I have some Celery tasks which are based on one abstract task.

class BaseTask(Task):
    abstract = True

    def on_failure(self, exc, task_id, args, kwargs, einfo):
        logging.info('Task failed')

    def on_success(self, retval, task_id, args, kwargs):
        logging.info('Task success')

    # NO SUCH METHOD IN TASK
    # def on_start(self):
    #     do_something_on_every_task_start()

@app.task(base=BaseTask)
def task1(x):
    print x

@app.task(base=BaseTask)
def task2(y):
    print y

I want to do something on start of every task execution. Are there any possibilities to achieve this result? Thanks.

Upvotes: 8

Views: 3554

Answers (2)

atv
atv

Reputation: 2000

If need 'semantically beautiful' method, you can try signals.

http://docs.celeryproject.org/en/latest/userguide/signals.html

Here is another useful post on pre/post task callbacks.

https://gist.github.com/nuria/35f332acfb84ecf80e3b

Upvotes: 2

Louis
Louis

Reputation: 151391

You can override the __call__ method. This will work both if you call the task synchronously with task1(1) or call it asynchronously, with task1.delay(1). In the following example, you'll get "CALLED!" on the console or in the logs depending on how you call the task and how your logging is setup.

class BaseTask(Task):
    abstract = True

    def __call__(self, *args, **kwargs):
        print "CALLED!"
        return super(BaseTask, self).__call__(*args, **kwargs)

@app.task(base=BaseTask)
def task1(x):
    print x

@app.task(base=BaseTask)
def task2(y):
    print y

Upvotes: 12

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