Python+Celery：链接作业？

发布于 2024-09-27 00:39:46 字数 808 浏览 2 评论 0原文

Celery 文档表明这是一个让任务等待其他任务的结果是个坏主意……但是建议的解决方案（参见“好”标题）还有一些不足之处。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）。

那么，有没有什么方法可以“链接”作业，以便调用者获得最终作业的结果呢？例如，使用 add 示例：

>>> add3 = add.subtask(args=(3, ))
>>> add.delay(1, 2, callback=add3).get()
6

或者，可以返回 Result 的实例吗？例如：

@task
def add(x, y, callback=None):
    result = x + y
    if callback:
        return subtask(callback).delay(result)
    return result

这将使链中“最终”作业的结果可以通过简单的操作来检索：

result = add(1, 2, callback=add3).delay()
while isinstance(result, Result):
    result = result.get()
print "result:", result

原文

The Celery documentation suggests 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).

So, is there any way of “chaining” jobs, so the caller gets the result of the final job? Eg, to use the add example:

>>> add3 = add.subtask(args=(3, ))
>>> add.delay(1, 2, callback=add3).get()
6

Alternately, is it OK to return instances of Result? For example:

@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:

result = add(1, 2, callback=add3).delay()
while isinstance(result, Result):
    result = result.get()
print "result:", result

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海拔太高太耀眼 2024-10-04 00:39:46

你可以用芹菜链来做。请参阅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

链接作业：

# 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()

You can do it with a celery chain. See 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:

# 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()

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