问题描述

import celery
def temptask(n):
    header=list(tempsubtask.si(i) for i in range(n))
    callback=templink.si('printed at last?')
    r = celery.chord(celery.group(header))(callback)
    return r

@task()
def tempsubtask(i):
    print i    
    for x in range(i):
        time.sleep(2)
        current_task.update_state(
            state='PROGRESS', meta={'completed': x, 'total': i })

@task()
def templink(x):
    print 'this should be run at last %s'%x

#executing temptask
r = temptask(100)

推荐答案

搜索了几个小时后，我偶然发现了http://www.manasupo.com/2012/03/chord-progress-in-celery.html。虽然那里的解决方案对我来说不是开箱即用的，但它确实激励了我尝试类似的东西。

from celery.utils import uuid
from celery import chord

class ProgressChord(chord):

    def __call__(self, body=None, **kwargs):
        _chord = self.type
        body = (body or self.kwargs['body']).clone()
        kwargs = dict(self.kwargs, body=body, **kwargs)
        if _chord.app.conf.CELERY_ALWAYS_EAGER:
            return self.apply((), kwargs)
        callback_id = body.options.setdefault('task_id', uuid())
        r= _chord(**kwargs)
        return _chord.AsyncResult(callback_id), r

我使用的不是celery.chord，而是ProgressChord，如下所示：

def temptask(n):
    header=list(tempsubtask.si(i) for i in range(n))
    callback=templink.si('printed at last?')
    r = celery.Progresschord(celery.group(header))(callback)
    return r

r的返回值包含同时具有回调异步结果和组结果的元组。所以成功看起来是这样的：

In [3]: r
Out[3]: 
(<AsyncResult: bf87507c-14cb-4ac4-8070-d32e4ff326a6>,
 <GroupResult: af69e131-5a93-492d-b985-267484651d95 [4672cbbb-8ec3-4a9e-971a-275807124fae, a236e55f-b312-485c-a816-499d39d7de41, e825a072-b23c-43f2-b920-350413fd5c9e, e3f8378d-fd02-4a34-934b-39a5a735871d, c4f7093b-9f1a-4e5e-b90d-66f83b9c97c4, d5c7dc2c-4e10-4e71-ba2b-055a33e15f02, 07b1c6f7-fe95-4c1f-b0ba-6bc82bceaa4e, 00966cb8-41c2-4e95-b5e7-d8604c000927, e039c78e-6647-4c8d-b59b-e9baf73171a0, 6cfdef0a-25a2-4905-a40e-fea9c7940044]>)

我继承并重写了[celery.chord][1]而不是[celery.task.chords.Chord][2]，因为我在任何地方都找不到它的源。

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