Question

我使用芹菜来更新我的新闻聚合网站中的RSS源。我为每个Feed使用了一个@task，事情看起来效果很好。

有一个细节，我不能确定处理得好：所有的Feed都是每分钟用@periodic_task更新一次，但是当一个新的启动时，如果一个Feed仍在从上一个周期性任务中更新怎么办？（例如，如果Feed非常慢，或者离线并且任务保持在重试循环中）

目前我存储任务结果并检查其状态如下：

import socket
from datetime import timedelta
from celery.decorators import task, periodic_task
from aggregator.models import Feed


_results = {}


@periodic_task(run_every=timedelta(minutes=1))
def fetch_articles():
    for feed in Feed.objects.all():
        if feed.pk in _results:
            if not _results[feed.pk].ready():
                # The task is not finished yet
                continue
        _results[feed.pk] = update_feed.delay(feed)


@task()
def update_feed(feed):
    try:
        feed.fetch_articles()
    except socket.error, exc:
        update_feed.retry(args=[feed], exc=exc)

也许有一种更复杂/更健壮的方法可以使用我错过的一些芹菜机制来达到相同的效果？

Answer 1

根据MattH的回答，您可以使用这样的装饰器：

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

然后，像这样使用它......

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

Answer 2

从官方文档：Ensuring a task is only executed one at a time。

Answer 3

使用https://pypi.python.org/pypi/celery_once似乎做得非常好，包括报告错误和针对某些参数进行独特性测试。

您可以执行以下操作：

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!"

只需要项目中的以下设置：

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

Answer 4

如果你正在寻找一个不使用Django的例子，那么try this example（警告：改为使用Redis，我已经在使用它）。

装饰器代码如下（完全归功于文章的作者，请阅读它）

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 5

我想知道为什么没有人提到使用 celery.app.control.inspect().active() 来获取当前正在运行的任务列表。不是实时的吗？因为否则它会很容易实现，例如：

def unique_task(callback,  *decorator_args, **decorator_kwargs):
    """
    Decorator to ensure only one instance of the task is running at once.
    """
    @wraps(callback)
    def _wrapper(celery_task, *args, **kwargs):
        active_queues = task.app.control.inspect().active()
        if active_queues:
            for queue in active_queues:
                for running_task in active_queues[queue]:
                    # Discard the currently running task from the list.
                    if task.name == running_task['name'] and task.request.id != running_task['id']:
                        return f'Task "{callback.__name__}()" cancelled! already running...'

        return callback(celery_task, *args, **kwargs)

    return _wrapper

然后只是将装饰器应用于相应的任务：

@celery.task(bind=True)
@unique_task
def my_task(self):
    # task executed once at a time.
    pass

Answer 6

芹菜在单一主机上工作的解决方案，其可靠性更高1.其他类型（没有像redis这样的依赖关系）基于锁的差异文件不适用于并发性更高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()

用芹菜运行“独特”的任务

6 个答案: