问题
I have celery beat and celery (four workers) to do some processing steps in bulk. One of those tasks is roughly along the lines of, "for each X that hasn't had a Y created, create a Y."
The task is run periodically at a semi-rapid rate (10sec). The task completes very quickly. There are other tasks going on as well.
I've run into the issue multiple times in which the beat tasks apparently become backlogged, and so the same task (from different beat times) are executed simultaneously, causing incorrectly duplicated work. It also appears that the tasks are executed out-of-order.
Is it possible to limit celery beat to ensure only one outstanding instance of a task at a time? Is setting something like
rate_limit=5on the task the "correct" way to of doing this?Is it possible to ensure that beat tasks are executed in-order, e.g. instead of dispatching a task, beat adds it to a task chain?
What's the best way of handling this, short of making those tasks themselves execute atomically and are safe to be executed concurrently? That was not a restriction I would have expected of beat tasks…
The task itself is defined naïvely:
@periodic_task(run_every=timedelta(seconds=10))
def add_y_to_xs():
# Do things in a database
return
Here's an actual (cleaned) log:
[00:00.000]foocorp.tasks.add_y_to_xs sent. id->#1[00:00.001]Received task: foocorp.tasks.add_y_to_xs[#1][00:10.009]foocorp.tasks.add_y_to_xs sent. id->#2[00:20.024]foocorp.tasks.add_y_to_xs sent. id->#3[00:26.747]Received task: foocorp.tasks.add_y_to_xs[#2][00:26.748]TaskPool: Apply #2[00:26.752]Received task: foocorp.tasks.add_y_to_xs[#3][00:26.769]Task accepted: foocorp.tasks.add_y_to_xs[#2] pid:26528[00:26.775]Task foocorp.tasks.add_y_to_xs[#2] succeeded in 0.0197986490093s: None[00:26.806]TaskPool: Apply #1[00:26.836]TaskPool: Apply #3[01:30.020]Task accepted: foocorp.tasks.add_y_to_xs[#1] pid:26526[01:30.053]Task accepted: foocorp.tasks.add_y_to_xs[#3] pid:26529[01:30.055]foocorp.tasks.add_y_to_xs[#1]: Adding Y for X id #9725[01:30.070]foocorp.tasks.add_y_to_xs[#3]: Adding Y for X id #9725[01:30.074]Task foocorp.tasks.add_y_to_xs[#1] succeeded in 0.0594762689434s: None[01:30.087]Task foocorp.tasks.add_y_to_xs[#3] succeeded in 0.0352867960464s: None
We're currently using Celery 3.1.4 with RabbitMQ as the transport.
EDIT Dan, here's what I came up with:
Dan, here's what I ended up using:
from sqlalchemy import func
from sqlalchemy.exc import DBAPIError
from contextlib import contextmanager
def _psql_advisory_lock_blocking(conn, lock_id, shared, timeout):
lock_fn = (func.pg_advisory_xact_lock_shared
if shared else
func.pg_advisory_xact_lock)
if timeout:
conn.execute(text('SET statement_timeout TO :timeout'),
timeout=timeout)
try:
conn.execute(select([lock_fn(lock_id)]))
except DBAPIError:
return False
return True
def _psql_advisory_lock_nonblocking(conn, lock_id, shared):
lock_fn = (func.pg_try_advisory_xact_lock_shared
if shared else
func.pg_try_advisory_xact_lock)
return conn.execute(select([lock_fn(lock_id)])).scalar()
class DatabaseLockFailed(Exception):
pass
@contextmanager
def db_lock(engine, name, shared=False, block=True, timeout=None):
"""
Context manager which acquires a PSQL advisory transaction lock with a
specified name.
"""
lock_id = hash(name)
with engine.begin() as conn, conn.begin():
if block:
locked = _psql_advisory_lock_blocking(conn, lock_id, shared,
timeout)
else:
locked = _psql_advisory_lock_nonblocking(conn, lock_id, shared)
if not locked:
raise DatabaseLockFailed()
yield
And the celery task decorator (used only for periodic tasks):
from functools import wraps
from preo.extensions import db
def locked(name=None, block=True, timeout='1s'):
"""
Using a PostgreSQL advisory transaction lock, only runs this task if the
lock is available. Otherwise logs a message and returns `None`.
"""
def with_task(fn):
lock_id = name or 'celery:{}.{}'.format(fn.__module__, fn.__name__)
@wraps(fn)
def f(*args, **kwargs):
try:
with db_lock(db.engine, name=lock_id, block=block,
timeout=timeout):
return fn(*args, **kwargs)
except DatabaseLockFailed:
logger.error('Failed to get lock.')
return None
return f
return with_task
回答1:
The only way to do this is implementing a locking strategy yourself:
Read under the section here for the reference.
Like with cron, the tasks may overlap if the first task does not complete before the next. If that is a concern you should use a locking strategy to ensure only one instance can run at a time (see for example Ensuring a task is only executed one at a time).
回答2:
def skip_if_running(f):
u"""
не запускает задачу с такими же параметрами если она уже в обработке
"""
task_name = u'%s.%s' % (f.__module__, f.__name__)
mylog.info(u'skip decorator for %s' % task_name)
@wraps(f)
def fun(self, *args, **kwargs):
try:
uargs = unicode(args)
ukwargs = unicode(kwargs)
i = clr_app.control.inspect()
workers = i.active()
for worker, tasks in workers.items():
for task in tasks:
if task_name == task['name'] and uargs == task['args'] and ukwargs == task['kwargs'] and self.request.id != task['id']:
mylog.warning(u'task %s (%s, %s) is started on %s, skip current' % (task_name, uargs, ukwargs, worker))
return None
except Exception as e:
mylog.error(e)
return f(*args, **kwargs)
return fun
@clr_app.task(bind=True)
@skip_if_running
def test_single_task(arg):
pass
回答3:
I solved the issue using celery-once which I extended to celery-one.
Both serve for your issue. It uses Redis to lock a running task. celery-one will also keep track of the task which is locking.
A very simple usage example for celery beat follows. In the code below, slow_task is scheduled every 1 second, but it's completion time is 5 seconds. Normal celery would schedule the task each second even if it is already running. celery-one would prevent this.
celery = Celery('test')
celery.conf.ONE_REDIS_URL = REDIS_URL
celery.conf.ONE_DEFAULT_TIMEOUT = 60 * 60
celery.conf.BROKER_URL = REDIS_URL
celery.conf.CELERY_RESULT_BACKEND = REDIS_URL
from datetime import timedelta
celery.conf.CELERYBEAT_SCHEDULE = {
'add-every-30-seconds': {
'task': 'tasks.slow_task',
'schedule': timedelta(seconds=1),
'args': (1,)
},
}
celery.conf.CELERY_TIMEZONE = 'UTC'
@celery.task(base=QueueOne, one_options={'fail': False})
def slow_task(a):
print("Running")
sleep(5)
return "Done " + str(a)
回答4:
I took a crack at writing a decorator to use Postgres advisory locking similar to what erydo alluded to in his comment.
It's not very pretty, but seems to work correctly. This is with SQLAlchemy 0.9.7 under Python 2.7.
from functools import wraps
from sqlalchemy import select, func
from my_db_module import Session # SQLAlchemy ORM scoped_session
def pg_locked(key):
def decorator(f):
@wraps(f)
def wrapped(*args, **kw):
session = db.Session()
try:
acquired, = session.execute(select([func.pg_try_advisory_lock(key)])).fetchone()
if acquired:
return f(*args, **kw)
finally:
if acquired:
session.execute(select([func.pg_advisory_unlock(key)]))
return wrapped
return decorator
@app.task
@pg_locked(0xdeadbeef)
def singleton_task():
# only 1x this task can run at a time
pass
(Would welcome any comments on ways to improve this!)
回答5:
A distributed locking system is required, for those Celery beat instances are essentially different processes which might be across different hosts.
Central coordinate systems such as ZooKeeper and etcd is suitable for implementation of distributed locking system.
I recommend using etcd, which is lightweight and fast. There are several implementations of lock over etcd, such as:
python-etcd-lock
来源：https://stackoverflow.com/questions/20894771/celery-beat-limit-to-single-task-instance-at-a-time