Celery / Django - How to programmatically see the status of workers

Question

I just installed Celery, and I want to create a simple status page showing the current number of employees and their status.

Is it possible? From a web search, the best I found was celery.current_app.control.inspect()

But, as far as I see, he does not say anything about the workers. (I use Kombu with SQS for the backend, if that matters)

Answer 1

The celery worker documentation explains the output of inspect commands.

By default, using celery.current_app.control.inspect() , an inspector object is returned that allows you to query the status of all workers. For example, if you are executing this code with two working workers with the names “adder” and “sleep”:

  i = celery.current_app.control.inspect() i.registered() 

calling i.registered() might return something like:

  { ' [e-mail] ': ['tasks.add'], ' [e-mail] ': ['tasks.sleeptask'], } 

In conclusion, the inspector methods are registered , active , scheduled , etc. will return a dictionary with the results classified by workers selected by calling celery.current_app.control.inspect() (if none of the workers are passed as arguments, all employees are implicitly selected).