用芹菜运行"独特"的任务

Question

我使用芹菜来更新我的新闻聚合网站中的RSS提要.我为每个feed使用一个@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,但是如果一个feed仍在从最后一个周期性任务中更新时新的一个怎么办？(例如,如果Feed非常慢,或者离线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实现相同的结果？

Answer 1

基于MattH的答案,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:确保任务一次只执行一个.

Answer 3

使用https://pypi.python.org/pypi/celery_once似乎非常好,包括报告错误和针对某些参数测试唯一性.

你可以这样做:

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!"

只需要在项目中进行以下设置:

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,那么试试这个例子(警告:使用Redis代替,我已经使用过).

装饰器代码如下(完全归功于文章的作者,去看看)

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的列表。难道不是实时的吗？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

然后只需将装饰器应用到相应的任务即可：

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