Running “unique” tasks with celery

Question

I use celery to update RSS feeds in my news aggregation site. I use one @task for each feed, and things seem to work nicely.

There\'s a detail that I\'m not sure to

Answer 1

Based on MattH's answer, you could use a decorator like this:

def single_instance_task(timeout):
    def task_exc(func):
        @functools.wraps(func)
        def wrapper(*args, **kwargs):
            lock_id = "celery-single-instance-" + func.__name__
            acquire_lock = lambda: cache.add(lock_id, "true", timeout)
            release_lock = lambda: cache.delete(lock_id)
            if acquire_lock():
                try:
                    func(*args, **kwargs)
                finally:
                    release_lock()
        return wrapper
    return task_exc

then, use it like so...

@periodic_task(run_every=timedelta(minutes=1))
@single_instance_task(60*10)
def fetch_articles()
    yada yada...

Answer 2

Using https://pypi.python.org/pypi/celery_once seems to do the job really nice, including reporting errors and testing against some parameters for uniqueness.

You can do things like:

from celery_once import QueueOnce
from myapp.celery import app
from time import sleep

@app.task(base=QueueOnce, once=dict(keys=('customer_id',)))
def start_billing(customer_id, year, month):
    sleep(30)
    return "Done!"

which just needs the following settings in your project:

ONCE_REDIS_URL = 'redis://localhost:6379/0'
ONCE_DEFAULT_TIMEOUT = 60 * 60  # remove lock after 1 hour in case it was stale

Answer 3

If you're looking for an example that doesn't use Django, then try this example (caveat: uses Redis instead, which I was already using).

The decorator code is as follows (full credit to the author of the article, go read it)

import redis

REDIS_CLIENT = redis.Redis()

def only_one(function=None, key="", timeout=None):
    """Enforce only one celery task at a time."""

    def _dec(run_func):
        """Decorator."""

        def _caller(*args, **kwargs):
            """Caller."""
            ret_value = None
            have_lock = False
            lock = REDIS_CLIENT.lock(key, timeout=timeout)
            try:
                have_lock = lock.acquire(blocking=False)
                if have_lock:
                    ret_value = run_func(*args, **kwargs)
            finally:
                if have_lock:
                    lock.release()

            return ret_value

        return _caller

    return _dec(function) if function is not None else _dec

Answer 4

This solution for celery working at single host with concurency greater 1. Other kinds (without dependencies like redis) of locks difference file-based don't work with concurrency greater 1.

class Lock(object):
    def __init__(self, filename):
        self.f = open(filename, 'w')

    def __enter__(self):
        try:
            flock(self.f.fileno(), LOCK_EX | LOCK_NB)
            return True
        except IOError:
            pass
        return False

    def __exit__(self, *args):
        self.f.close()


class SinglePeriodicTask(PeriodicTask):
    abstract = True
    run_every = timedelta(seconds=1)

    def __call__(self, *args, **kwargs):
        lock_filename = join('/tmp',
                             md5(self.name).hexdigest())
        with Lock(lock_filename) as is_locked:
            if is_locked:
                super(SinglePeriodicTask, self).__call__(*args, **kwargs)
            else:
                print 'already working'


class SearchTask(SinglePeriodicTask):
    restart_delay = timedelta(seconds=60)

    def run(self, *args, **kwargs):
        print self.name, 'start', datetime.now()
        sleep(5)
        print self.name, 'end', datetime.now()

Answer 5

From the official documentation: Ensuring a task is only executed one at a time.