How to make a celery task fail from within the tas

Question

Under some conditions, I want to make a celery task fail from within that task. I tried the following:

from celery.task import task
from celery import states

@task()
def run_simulation():
    if some_condition:
        run_simulation.update_state(state=states.FAILURE)
        return False

However, the task still reports to have succeeded:

Task sim.tasks.run_simulation[9235e3a7-c6d2-4219-bbc7-acf65c816e65] succeeded in 1.17847704887s: False

It seems that the state can only be modified while the task is running and once it is completed - celery changes the state to whatever it deems is the outcome (refer to this question). Is there any way, without failing the task by raising an exception, to make celery return that the task has failed?

Answer 1

I'd like to further expand on Pierre's answer as I've encountered some issues using the suggested solution.

To allow custom fields when updating a task's state to states.FAILURE, it is important to also mock some attributes that a FAILURE state would have (notice exc_type and exc_message) While the solution will terminate the task, any attempt to query the state (For example - to fetch the 'REASON FOR FAILURE' value) will fail.

Below is a snippet for reference I took from: https://www.distributedpython.com/2018/09/28/celery-task-states/

@app.task(bind=True)
def task(self):
    try:
        raise ValueError('Some error')
    except Exception as ex:
        self.update_state(
            state=states.FAILURE,
            meta={
                'exc_type': type(ex).__name__,
                'exc_message': traceback.format_exc().split('\n')
                'custom': '...'
            })
        raise Ignore()

Answer 2

I got an interesting reply on this question from Ask Solem, where he proposes an 'after_return' handler to solve the issue. This might be an interesting option for the future.

In the meantime I solved the issue by simply returning a string 'FAILURE' from the task when I want to make it fail and then checking for that as follows:

result = AsyncResult(task_id)
if result.state == 'FAILURE' or (result.state == 'SUCCESS' and result.get() == 'FAILURE'):
    # Failure processing task 

Answer 3

To mark a task as failed without raising an exception, update the task state to FAILURE and then raise an Ignore exception, because returning any value will record the task as successful, an example:

from celery import Celery, states
from celery.exceptions import Ignore

app = Celery('tasks', broker='amqp://guest@localhost//')

@app.task(bind=True)
def run_simulation(self):
    if some_condition:
        # manually update the task state
        self.update_state(
            state = states.FAILURE,
            meta = 'REASON FOR FAILURE'
        )

        # ignore the task so no other state is recorded
        raise Ignore()

But the best way is to raise an exception from your task, you can create a custom exception to track these failures:

class TaskFailure(Exception):
   pass

And raise this exception from your task:

if some_condition:
    raise TaskFailure('Failure reason')