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用芹菜执行“独特”任务

celery django python

Luper Rouch François Lagunas · 技术社区 · 14 年前

我用芹菜更新我的新闻聚合网站的RSS订阅源。我为每个feed使用一个@task,事情看起来运行得很好。

当前我存储任务结果并检查其状态,如下所示:

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)

5 回复 | 直到 7 年前

mmoya Mohammed Azharuddin Shaikh 11 年前

从官方文件来看: Ensuring a task is only executed one at a time .

Peter Kilczuk 12 年前

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...

vdboor 9 年前

使用 https://pypi.python.org/pypi/celery_once 这项工作似乎做得很好,包括报告错误和测试一些参数的唯一性。

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

jbkkd 9 年前

如果你想找一个不使用Django的例子,那么 try this example (注意:改用Redis,我已经用过了)。

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

user12397901 11 年前

此解决方案适用于芹菜在同一主机上工作时并发性大于1。其他类型的锁(没有依赖项,比如redis)基于差异文件的锁在并发性大于1时不起作用。

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()