Python+Celery：链接工作？
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Python+Celery: Chaining jobs?
提问by David Wolever
The Celery documentationsuggests that it's a bad idea to have tasks wait on the results of other tasks… But the suggested solution (see “good” heading) leaves a something to be desired. Specifically, there's no clear way of getting the subtask's result back to the caller (also, it's kind of ugly).
将芹菜文件表明，这是一个坏主意，有任务等待的其他任务的结果。但建议的解决方案（见“好”的标题）离开是可喜爱的东西。具体来说，没有明确的方法将子任务的结果返回给调用者（而且，这有点难看）。
So, is there any way of “chaining” jobs, so the caller gets the result of the final job? Eg, to use the addexample:
那么，有没有办法“链接”作业，让调用者得到最终作业的结果？例如，使用add示例：
>>> add3 = add.subtask(args=(3, ))
>>> add.delay(1, 2, callback=add3).get()
6
Alternately, is it OK to return instances of Result? For example:
或者，是否可以返回 Result 的实例？例如：
@task
def add(x, y, callback=None):
result = x + y
if callback:
return subtask(callback).delay(result)
return result
This would let the result of the “final” job in the chain could be retrived with a simple:
这将使链中“最终”作业的结果可以通过一个简单的方式检索：
result = add(1, 2, callback=add3).delay()
while isinstance(result, Result):
result = result.get()
print "result:", result
采纳答案by ax003d
You can do it with a celery chain. See https://celery.readthedocs.org/en/latest/userguide/canvas.html#chains
你可以用芹菜链来做到这一点。见https://celery.readthedocs.org/en/latest/userguide/canvas.html#chains
@task()
def add(a, b):
time.sleep(5) # simulate long time processing
return a + b
Chaining job:
链式作业：
# import chain from celery import chain
# the result of the first add job will be
# the first argument of the second add job
ret = chain(add.s(1, 2), add.s(3)).apply_async()
# another way to express a chain using pipes
ret2 = (add.s(1, 2) | add.s(3)).apply_async()
...
# check ret status to get result
if ret.status == u'SUCCESS':
print "result:", ret.get()
