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?
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')