Celery First Steps - result.get()上的超时错误

Question

我在这里关注Celery First Steps教程:http://celery.readthedocs.org/en/latest/getting-started/first-steps-with-celery.html#keeping-results

我正在使用RabbitMQ继续学习本教程.

当我在做result.get(timeout = 1)时,它显示一个超时错误,即使它是一个简单的添加操作,我可以看到工作者在另一个窗口中运行并产生正确的结果(8)

(venv) C:\Volt\celerytest>ipython
Python 2.7.6 (default, Nov 10 2013, 19:24:18) [MSC v.1500 32 bit (Intel)]
Type "copyright", "credits" or "license" for more information.

IPython 2.1.0 -- An enhanced Interactive Python.
?         -> Introduction and overview of IPython's features.
%quickref -> Quick reference.
help      -> Python's own help system.
object?   -> Details about 'object', use 'object??' for extra details.

In [1]: from tasks import add

In [2]: a = add(1,3)

In [3]: a
Out[3]: 4

In [4]: a = add.delay(1,3)

In [5]: a.ready()
Out[5]: False

In [6]: a = add.delay(4,4)

In [7]: a.get(timeout=0.5)
---------------------------------------------------------------------------
TimeoutError                              Traceback (most recent call last)
<ipython-input-7-2c407a92720e> in <module>()
----> 1 a.get(timeout=0.5)

C:\Users\Som\Envs\venv\lib\site-packages\celery\result.pyc in get(self, timeout,
 propagate, interval, no_ack, follow_parents)
    167                 interval=interval,
    168                 on_interval=on_interval,
--> 169                 no_ack=no_ack,
    170             )
    171         finally:

C:\Users\Som\Envs\venv\lib\site-packages\celery\backends\amqp.pyc in wait_for(se
lf, task_id, timeout, cache, propagate, no_ack, on_interval, READY_STATES, PROPA
GATE_STATES, **kwargs)
    155                                     on_interval=on_interval)
    156             except socket.timeout:
--> 157                 raise TimeoutError('The operation timed out.')
    158
    159         if meta['status'] in PROPAGATE_STATES and propagate:

TimeoutError: The operation timed out.

In [8]:

tasks.py文件

from celery import Celery

app = Celery('tasks', backend='amqp', broker='amqp://')


@app.task
def add(x, y):
    return x + y

工人日志

[tasks]
  . tasks.add

[2014-07-17 13:00:33,196: INFO/MainProcess] Connected to amqp://guest:**@127.0.0
.1:5672//
[2014-07-17 13:00:33,211: INFO/MainProcess] mingle: searching for neighbors
[2014-07-17 13:00:34,220: INFO/MainProcess] mingle: all alone
[2014-07-17 13:00:34,240: WARNING/MainProcess] celery@SomsPC ready.
[2014-07-17 13:00:34,242: INFO/MainProcess] Received task: tasks.add[85ff75d8-38
b5-442a-a574-c8b976a33739]
[2014-07-17 13:00:34,243: INFO/MainProcess] Task tasks.add[85ff75d8-38b5-442a-a5
74-c8b976a33739] succeeded in 0.000999927520752s: 4
[2014-07-17 13:00:46,582: INFO/MainProcess] Received task: tasks.add[49de7c6b-96
72-485d-926e-a4e564ccc89a]
[2014-07-17 13:00:46,588: INFO/MainProcess] Task tasks.add[49de7c6b-9672-485d-92
6e-a4e564ccc89a] succeeded in 0.00600004196167s: 8

Answer 1

经过'芹菜第一步'后,我遇到了完全相同的问题.

我认为建议这样做的原因backend='amqp'.

对我有用的设置如下:

app = Celery('tasks', broker='amqp://guest@localhost//')
app.conf.CELERY_RESULT_BACKEND = 'db+sqlite:///results.sqlite'

根据文档,当使用AMQP结果后端时,每个结果只能被检索一次(它实际上是查询中的单个消息).

我想,你的工作进程检索它以便将结果打印到控制台:

Task tasks.add[49de7c6b-9672-485d-926e-a4e564ccc89a] succeeded in 0.00600004196167s: 8

所以你无法再次检索相同的结果.