如果芹菜不起作用，则返回正常功能

Question

我需要的是一个简单的黑客同步运行功能，如果芹菜不活跃。你知道吗

我尝试的是： is_celery_working返回False，尽管芹菜和Redis都在运行（分别是rancelery -A project worker -l debug和redis-server）。而且get_celery_worker_status总是给error状态。你知道吗

我在用芹菜配姜果。你知道吗

from project.celery import app

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


def sync_async(func):
    if is_celery_working():
        return func.delay
    else:
        return func

sync_async(some_func)(**its_args, **its_kwrgs)

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

def sync_async(func):
    status = get_celery_worker_status()
    if 'error' not in status:
        return func.delay
    else:
        return func

sync_async(some_func)(**its_args, **its_kwrgs)

Answer 1

简单的is_celery_working函数看起来是正确的。如果得到False，则可能需要使用可选的timeout参数将超时时间增加到5或10秒。你知道吗

def is_celery_working():
    result = app.control.broadcast('ping', reply=True, limit=1, timeout=5.0)
    return bool(result)  # True if at least one result

Answer 2

def sync_async(func, *args, **kwargs):
    try:
        func.delay(*args, **kwargs)
    except Exception as error:
        print('Celery not active', error)
        func(*args, **kwargs)

如果Redis服务器不工作，这只会给出一个错误。这对我来说很好，因为我假设如果Redis不工作，那么芹菜就停止了。你知道吗