Detect whether Celery is Available/Running

Question

I'm using Celery to manage asynchronous tasks. Occasionally, however, the celery process goes down which causes none of the tasks to get executed. I would like to be able to check the status of celery and make sure everything is working fine, and if I detect any problems display an error message to the user. From the Celery Worker documentation it looks like I might be able to use ping or inspect for this, but ping feels hacky and it's not clear exactly how inspect is meant to be used (if inspect().registered() is empty?).

Any guidance on this would be appreciated. Basically what I'm looking for is a method like so:

def celery_is_alive():
    from celery.task.control import inspect
    return bool(inspect().registered()) # is this right??

EDIT: It doesn't even look like registered() is available on celery 2.3.3 (even though the 2.1 docs list it). Maybe ping is the right answer.

EDIT: Ping also doesn't appear to do what I thought it would do, so still not sure the answer here.

Solution 1

Here's the code I've been using. celery.task.control.Inspect.stats() returns a dict containing lots of details about the currently available workers, None if there are no workers running, or raises an IOError if it can't connect to the message broker. I'm using RabbitMQ - it's possible that other messaging systems might behave slightly differently. This worked in Celery 2.3.x and 2.4.x; I'm not sure how far back it goes.

def get_celery_worker_status():
    ERROR_KEY = "ERROR"
    try:
        from celery.task.control import inspect
        insp = inspect()
        d = insp.stats()
        if not d:
            d = { ERROR_KEY: 'No running Celery workers were found.' }
    except IOError as e:
        from errno import errorcode
        msg = "Error connecting to the backend: " + str(e)
        if len(e.args) > 0 and errorcode.get(e.args[0]) == 'ECONNREFUSED':
            msg += ' Check that the RabbitMQ server is running.'
        d = { ERROR_KEY: msg }
    except ImportError as e:
        d = { ERROR_KEY: str(e)}
    return d

Solution 2

From the documentation of celery 4.2:

from your_celery_app import app


def get_celery_worker_status():
    i = app.control.inspect()
    availability = i.ping()
    stats = i.stats()
    registered_tasks = i.registered()
    active_tasks = i.active()
    scheduled_tasks = i.scheduled()
    result = {
        'availability': availability,
        'stats': stats,
        'registered_tasks': registered_tasks,
        'active_tasks': active_tasks,
        'scheduled_tasks': scheduled_tasks
    }
    return result

of course you could/should improve the code with error handling...

Solution 3

To check the same using command line in case celery is running as daemon,

• Activate virtualenv and go to the dir where the 'app' is
• Now run : celery -A [app_name] status
• It will show if celery is up or not plus no. of nodes online

Source: http://michal.karzynski.pl/blog/2014/05/18/setting-up-an-asynchronous-task-queue-for-django-using-celery-redis/

Solution 4

The following worked for me:

import socket
from kombu import Connection

celery_broker_url = "amqp://localhost"

try:
    conn = Connection(celery_broker_url)
    conn.ensure_connection(max_retries=3)
except socket.error:
    raise RuntimeError("Failed to connect to RabbitMQ instance at {}".format(celery_broker_url))

Solution 5

One method to test if any worker is responding is to send out a 'ping' broadcast and return with a successful result on the first response.

from .celery import app  # the celery 'app' created in your project

def is_celery_working():
    result = app.control.broadcast('ping', reply=True, limit=1)
    return bool(result)  # True if at least one result

This broadcasts a 'ping' and will wait up to one second for responses. As soon as the first response comes in, it will return a result. If you want a False result faster, you can add a timeout argument to reduce how long it waits before giving up.

View more solutions