如果我以前不知道执行了哪个任务,如何提取任务结果? 这是设置: 给出以下来源('tasks.py'):
from celery import Celery
app = Celery('tasks', backend="db+mysql://u:p@localhost/db", broker = 'amqp://guest:guest@localhost:5672//')
@app.task
def add(x,y):
return x + y
@app.task
def mul(x,y):
return x * y
RabbitMQ 3.3.2在本地运行:
marcs-mbp:sbin marcstreeter$ ./rabbitmq-server
RabbitMQ 3.3.2. Copyright (C) 2007-2014 GoPivotal, Inc.
## ## Licensed under the MPL. See http://www.rabbitmq.com/
## ##
########## Logs: /usr/local/var/log/rabbitmq/rabbit@localhost.log
###### ## /usr/local/var/log/rabbitmq/rabbit@localhost-sasl.log
##########
Starting broker... completed with 10 plugins.
Celery 3.1.12在本地运行:
-------------- celery@Marcs-MacBook-Pro.local v3.1.12 (Cipater)
---- **** -----
--- * *** * -- Darwin-13.2.0-x86_64-i386-64bit
-- * - **** ---
- ** ---------- [config]
- ** ---------- .> app: tasks:0x105dea3d0
- ** ---------- .> transport: amqp://guest:**@localhost:5672//
- ** ---------- .> results: disabled
- *** --- * --- .> concurrency: 8 (prefork)
-- ******* ----
--- ***** ----- [queues]
-------------- .> celery exchange=celery(direct) key=celery
然后我可以导入方法并使用'task_id'检索结果:
from tasks import add, mul
from celery.result import AsyncResult
result = add.delay(2,2)
task_id = result.task_id
result.get() # 4
result = AsyncResult(id=task_id)
result.get() # 4
result = add.AsyncResult(id=task_id)
result.get() # 4
# and the same for the 'mul' task. Just imagine I put it here
在下一个例子中,我在进程之间拆分了这些步骤。在一个过程中,我像这样检索'task_id':
from tasks import add
result = add.delay(5,5)
task_id = result.task_id
在另一个过程中,如果我使用相同的'task_id'(复制并粘贴到另一个REPL,或在不同的HTTP请求中),如下所示:
from celery.result import AsyncResult
result = AsyncResult(id="copied_task_id", backend="db+mysql://u:p@localhost/db")
result.get() # AttributeError: 'str' object has no attribute 'get_task_meta'
result.state # AttributeError: 'str' object has no attribute 'get_task_meta'
result.status # AttributeError: 'str' object has no attribute 'get_task_meta'
如果我这样做又在另一个过程中:
from task import add # in this instance I know that an add task was performed
result = add.AsyncResult(id="copied_task_id")
result.status # "SUCCESSFUL"
result.state # "SUCCESSFUL"
result.get() # 10
我希望能够在不知道什么任务产生结果的情况下获得结果。在我的真实环境中,我计划将此task_id返回给客户端,让他们通过HTTP请求查询其工作状态。
答案 0 :(得分:16)
好的 - 所以我一直在寻找解决方案很长一段时间,现在我终于正式发布了这个并查看documentation我发现的this gem:
类celery.result.AsyncResult (id,backend = None,task_name = None, app =无,父=无)
查询任务状态。
参数:
id - 请参阅id。
后端 - 请参阅backend。
异常 TimeoutError
超时引发错误。
AsyncResult.app =无
因此,我没有提供后端参数,而是提供了“app”参数,而不是这样:
from celery.result import AsyncResult
from task import app
# Assuming add.delay(10,10) was called in another process
# and that I'm using a 'task_id' I retrieved from that process
result = AsyncResult(id='copied_task_id', app=app)
result.state # 'SUCCESSFUL'
result.get() # 20
这对许多人来说可能是显而易见的。这不是我。现在我可以说的是,这个解决方案“只是有效”,但如果我知道这是获得批准的方式,我会感到更舒服。如果您知道文档中的一个部分使其更清晰,请将其发布在评论中或作为答案,如果可以,我会选择它作为答案。
答案 1 :(得分:0)
万一它对任何人都有用,那么事实证明backend
参数不需要字符串,而是后端对象:How do I override the backend for celery tasks
对我有用的是:
from celery.backends.rpc import RPCBackend
from myapp.workers.main import app as worker
@worker.task(backend=RPCBackend(app=worker))
def status_check():
return "OK"