我如何删除所有挂起的任务,而不知道每个任务的task_id ?


当前回答

如果你想删除所有挂起的任务,以及活动和保留的任务,以完全停止Celery,这对我来说是有效的:

from proj.celery import app
from celery.task.control import inspect, revoke

# remove pending tasks
app.control.purge()

# remove active tasks
i = inspect()
jobs = i.active()
for hostname in jobs:
    tasks = jobs[hostname]
    for task in tasks:
        revoke(task['id'], terminate=True)

# remove reserved tasks
jobs = i.reserved()
for hostname in jobs:
    tasks = jobs[hostname]
    for task in tasks:
        revoke(task['id'], terminate=True)

其他回答

从文档中可以看出:

$ celery -A proj purge

or

from proj.celery import app
app.control.purge()

(编辑:更新与当前的方法。)

芹菜4 + 芹菜清除命令清除所有已配置的任务队列

celery -A *APPNAME* purge

编程:

from proj.celery import app
app.control.purge()

所有挂起的任务将被清除。 参考:celerydoc

芹菜3+:

CLI:

$ celery -A proj purge

编程:

>>> from proj.celery import app
>>> app.control.purge()

http://docs.celeryproject.org/en/latest/faq.html#how-do-i-purge-all-waiting-tasks

如果你想删除所有挂起的任务,以及活动和保留的任务,以完全停止Celery,这对我来说是有效的:

from proj.celery import app
from celery.task.control import inspect, revoke

# remove pending tasks
app.control.purge()

# remove active tasks
i = inspect()
jobs = i.active()
for hostname in jobs:
    tasks = jobs[hostname]
    for task in tasks:
        revoke(task['id'], terminate=True)

# remove reserved tasks
jobs = i.reserved()
for hostname in jobs:
    tasks = jobs[hostname]
    for task in tasks:
        revoke(task['id'], terminate=True)

对于芹菜5.0+,从CLI执行该操作,并针对特定的队列:

—APP_NAME清除——QUEUE_NAME

如果你想像我一样在一个步骤中做,那么在最后添加-f选项来跳过确认步骤。