Question

我有这样的任务

@celery.task(bind=True, max_retries = 5)
def send_notification_task(self, user_id, content):
    print 'Sending message to pubnub'

    msg = prepare_message(16182 , 16182, 'Android', 'This is a text message')  

    def _callback(message):
        print 'success callback'  
        print(message)

    def _error(message):
        print 'error callback'  
        print self.request.retries
        print self
        countdown = int(random.uniform(2, 4) ** self.request.retries)
        print countdown
        raise self.retry(countdown=countdown)

    publish('16182', msg, _callback, _error)

    print 'returning success to celery'

    current_task.update_state(state='PROGRESS', meta={'description': 'Doing some task', 'current': 59, 'tota': 73})

在这里，我想仅在回调方法即成功或失败时将任务标记为成功/失败。但是，只要此函数返回，任务就会被标记为成功（发布本身就是异步调用。）

我该如何处理？

我已经尝试将任务标记为PROGRESS作为最后一行。

Answer 1

在这里回答我自己的问题。我最终使用的是自定义CountDownLatch。这是更新的代码：

def send_notification_task(self, user, content):
    a = datetime.now()

    latch = CountDownLatch(1)
    print 'Sending message to pubnub'

    status = {'result' : 0}

    print "Here is the content:"
    print content

    if 'image_url' in content:
        image_url = content['image_url']
    else:
        image_url = None

    if 'category' in content:
        category = content['category']
    else:
        category = 'pnr'

    msg = prepare_message(user, content['title'], content['message'], image_url, category)

    if msg==None:
        print 'Not sending None message (probably iOS device)'
        return

    def _callback(message):
        print 'success callback'  
        print(message)
        status['result'] = 1
        latch.count_down()


    def _error(message):
        print 'error callback'
        countdown = exponential_backoff(self.request.retries)
        status['result'] = 2
        print str(status) + ' error'
        latch.count_down()

    print msg
    print user['user_channel_id']     

    publish(user['user_channel_id'], msg, _callback, _error)

    latch.await()

    # while status['result'] == 0:
    #     print 'loop'
    #     print status
    #     time.sleep(.001)

    if status['result'] == 2:
        print 'retry'
        countdown = exponential_backoff(self.request.retries)
        raise self.retry(countdown = countdown)
    elif status['result'] == 1:
        print 'success' 
        b = datetime.now()

        c = b - a

        print 'send_notification_task {0}'.format(c)
        return 'success'

倒计时，

import threading

class CountDownLatch(object):
    def __init__(self, count=1):
        self.count = count
        self.lock = threading.Condition()

    def count_down(self):
        self.lock.acquire()
        self.count -= 1
        if self.count <= 0:
            self.lock.notifyAll()
        self.lock.release()

    def await(self):
        self.lock.acquire()
        while self.count > 0:
            self.lock.wait()
        self.lock.release()

如何防止任务标记在返回时成功[芹菜]

1 个答案: