Question

我有这个Celery任务：

@app.task
def do_something(with_this):

    # instantiate a class from a third party library
    instance = SomeClass()

    # this class uses callbacks to send progress info about
    # the status and progress of what we're doing
    def progress_callback(data):
        # this status will change to 'finished' later
        # but the return value that I want as the task result won't be returned
        # so this is where I should mark the task as done manually
        if data['status'] == 'working':
            # I create a custom state for this task
            do_something.update_state(
                state = 'PROGRESS',
                meta = data['progress']
            )

    # adding the callback to the instance
    instance.add_callback(progress_callback)

    # use the instance to do what I want
    # this functions returns a value that I don't want as the task result
    # so leaving this function without a return statement will make it None
    instance.do_job(with_this)

如何将任务标记为手动完成？

在这种情况下，函数到达结尾而没有任何return语句，因此我得到的task.result是None，我想将传递给回调函数的数据设置为结果，将任务标记为已完成。

我尝试使用：

app.backend.mark_as_done(do_something.request.id, data)

它成功设置了任务的状态和结果，但稍后将结果设置为函数的返回值None。

Answer 1

我终于找到了存储任务状态和结果的解决方案，然后通过引发Ignore异常来忽略任务，例如：

from celery.exceptions import Ignore

@app.task
def do_something(with_this):

    # store the state and result manually
    # the SUCCESS state is set by this method
    app.backend.mark_as_done(
        do_something.request.id,
        the_data_to_store
    )

    # we can also use update_state which calls
    # backend.store_result just like mark_as_done
    # but we have to set the state in this case
    do_something.update_state(
        state = celery.states.SUCCESS,
        meta = the_data_to_store
    )

    # ignore the task so no other state is recorded
    # like what was happening with my function in the question
    # the task will still be acknowledged
    raise Ignore()

如果您无法返回要存储的数据，这将非常有用。

如何手动将Celery任务标记为已完成并设置其结果？

1 个答案: