python - celery - 链接组和子任务。 -> 乱序执行

Question

当我遇到以下情况时

group1 = group(task1.si(), task1.si(), task1.si())
group2 = group(task2.si(), task2.si(), task2.si())

workflow = chain(group1, group2, task3.si())

直观的解释是 task3 应该只在组 2 中的所有任务都完成后执行。

实际上，任务 3 在 group1 已启动但尚未完成时执行。

我做错了什么？

Answer 1

事实证明，在 celery 中，您不能将两组链接在一起。
我怀疑这是因为与任务链接的组自动成为一个和弦
--> celery 文档:http://docs.celeryproject.org/en/latest/userguide/canvas.html

Chaining a group together with another task will automatically upgrade it to be a chord:

组返回父任务。当将两组链接在一起时，我怀疑当第一组完成时，和弦开始回调“任务”。我怀疑这个“任务”实际上是第二组的“父任务”。我进一步怀疑这个父任务在完成启动组内的所有子任务后立即完成，因此执行第二组之后的下一个项目。

为了证明这一点，这里有一些示例代码。您需要已经有一个正在运行的 celery 实例。

# celery_experiment.py

from celery import task, group, chain, chord
from celery.signals import task_sent, task_postrun, task_prerun

import time
import logging

import random
random.seed()

logging.basicConfig(level=logging.DEBUG)

### HANDLERS ###    
@task_prerun.connect()
def task_starting_handler(sender=None, task_id=None, task=None, args=None, kwargs=None, **kwds):    
    try:
        logging.info('[%s] starting' % kwargs['id'])
    except KeyError:
        pass

@task_postrun.connect()
def task_finished_handler(sender=None, task_id=None, task=None, args=None, kwargs=None, retval=None, state=None, **kwds):
    try:    
        logging.info('[%s] finished' % kwargs['id'])
    except KeyError:
        pass


def random_sleep(id):
    slp = random.randint(1, 3)
    logging.info('[%s] sleep for %ssecs' % (id, slp))
    time.sleep(slp)

@task()
def thing(id):
    logging.info('[%s] begin' % id)
    random_sleep(id)
    logging.info('[%s] end' % id)


def exec_exp():
    st = thing.si(id='st')
    st_arr = [thing.si(id='st_arr1_a'), thing.si(id='st_arr1_b'), thing.si(id='st_arr1_c'),]
    st_arr2 = [thing.si(id='st_arr2_a'), thing.si(id='st_arr2_b'),]
    st2 = thing.si(id='st2')
    st3 = thing.si(id='st3')
    st4 = thing.si(id='st4')

    grp1 = group(st_arr)
    grp2 = group(st_arr2)

    # chn can chain two groups together because they are seperated by a single subtask
    chn = (st | grp1 | st2 | grp2 | st3 | st4)

    # in chn2 you can't chain two groups together. what will happen is st3 will start before grp2 finishes
    #chn2 = (st | st2 | grp1 | grp2 | st3 |  st4)

    r = chn()
    #r2 = chn2()