Question

我使用Django 1.6.11和RabbitMQ 3.5.4运行Celery 3.1.18，并尝试在失败状态下测试我的异步任务（CELERY_ALWAYS_EAGER = True）。但是，我无法在错误回调中获得正确的“结果”。 Celery docs中的示例显示：

@app.task(bind=True)
def error_handler(self, uuid):
    result = self.app.AsyncResult(uuid)
    print('Task {0} raised exception: {1!r}\n{2!r}'.format(
          uuid, result.result, result.traceback))

当我这样做时，我的结果仍然是“待命”，result.result = ''和result.traceback=''。但是我的.apply_async调用返回的实际结果具有正确的“FAILURE”状态和回溯。

我的代码（基本上是一个Django Rest Framework RESTful端点，它解析.tar.gz文件，然后在文件解析完成后向用户发送通知）：

views.py：

from producer_main.celery import app as celery_app

@celery_app.task()
def _upload_error_simple(uuid):
    print uuid
    result = celery_app.AsyncResult(uuid)
    print result.backend
    print result.state
    print result.result
    print result.traceback
    msg = 'Task {0} raised exception: {1!r}\n{2!r}'.format(uuid,
                                                           result.result,
                                                           result.traceback)


class UploadNewFile(APIView):
    def post(self, request, repository_id, format=None):
        try:    
            uploaded_file = self.data['files'][self.data['files'].keys()[0]]
            self.path = default_storage.save('{0}/{1}'.format(settings.MEDIA_ROOT,
                                                              uploaded_file.name),
                                             uploaded_file)
            print type(import_file)
            self.async_result = import_file.apply_async((self.path,  request.user),
                                                        link_error=_upload_error_simple.s())


            print 'results from self.async_result:'
            print self.async_result.id
            print self.async_result.backend
            print self.async_result.state
            print self.async_result.result
            print self.async_result.traceback
            return Response()
        except (PermissionDenied, InvalidArgument, NotFound, KeyError) as ex:
            gutils.handle_exceptions(ex)

tasks.py：

from producer_main.celery import app
from utilities.general import upload_class


@app.task
def import_file(path, user):
    """Asynchronously import a course."""
    upload_class(path, user)

celery.py：

"""
As described in
http://celery.readthedocs.org/en/latest/django/first-steps-with-django.html
"""
from __future__ import absolute_import

import os
import logging

from celery import Celery

os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'producer_main.settings')

from django.conf import settings

log = logging.getLogger(__name__)

app = Celery('producer')  # pylint: disable=invalid-name

# Using a string here means the worker will not have to
# pickle the object when using Windows.
app.config_from_object('django.conf:settings')
app.autodiscover_tasks(lambda: settings.INSTALLED_APPS)  # pragma: no cover

@app.task(bind=True)
def debug_task(self):
    print('Request: {0!r}'.format(self.request))

我的后端配置如下：

CELERY_ALWAYS_EAGER = True
CELERY_EAGER_PROPAGATES_EXCEPTIONS = False
BROKER_URL = 'amqp://'
CELERY_RESULT_BACKEND = 'redis://localhost'
CELERY_RESULT_PERSISTENT = True
CELERY_IGNORE_RESULT = False

当我为link_error状态运行unittest时，我得到：

Creating test database for alias 'default'...
<class 'celery.local.PromiseProxy'>
130ccf13-c2a0-4bde-8d49-e17eeb1b0115
<celery.backends.redis.RedisBackend object at 0x10aa2e110>
PENDING
None
None
results from self.async_result:
130ccf13-c2a0-4bde-8d49-e17eeb1b0115
None
FAILURE
Non .zip / .tar.gz file passed in.
Traceback (most recent call last):

因此我的_upload_error_simple()方法无法使用任务结果，但可以从self.async_result返回的变量中找到它们。

Answer 1

我无法使link和link_error回调起作用，因此我最终不得不使用the docs中描述的on_failure和on_success任务方法和this SO question。我的tasks.py看起来像是：

class ErrorHandlingTask(Task):
    abstract = True

    def on_failure(self, exc, task_id, targs, tkwargs, einfo):
        msg = 'Import of {0} raised exception: {1!r}'.format(targs[0].split('/')[-1],
                                                             str(exc))

    def on_success(self, retval, task_id, targs, tkwargs):
        msg = "Upload successful. You may now view your course."    

@app.task(base=ErrorHandlingTask)
def import_file(path, user):
    """Asynchronously import a course."""
    upload_class(path, user)

Answer 2

您似乎将_upload_error()作为您班级的绑定方法 - 这可能不是您想要的。尝试将其作为一项独立的任务：

@celery_app.task(bind=True)
def _upload_error(self, uuid):
    result = celery_app.AsyncResult(uuid)
    msg = 'Task {0} raised exception: {1!r}\n{2!r}'.format(uuid,
                                                       result.result,
                                                       result.traceback)

class Whatever(object):
    ....
    self.async_result = import_file.apply_async((self.path, request.user),
                                                link=self._upload_success.s(
                                                    "Upload finished."),
                                                link_error=_upload_error.s())

实际上不需要self参数，因为它没有被使用，所以你可以这样做：

@celery_app.task()
def _upload_error(uuid):
    result = celery_app.AsyncResult(uuid)
    msg = 'Task {0} raised exception: {1!r}\n{2!r}'.format(uuid,
                                                       result.result,
                                                       result.traceback)

请注意缺少bind=True和self

Answer 3

请注意UUID实例！

如果您尝试获取ID为string类型而不是UUID类型的任务的状态，则您只会获得PENDING状态。

from uuid import UUID
from celery.result import AsyncResult

task_id = UUID('d4337c01-4402-48e9-9e9c-6e9919d5e282')

print(AsyncResult(task_id).state)
# PENDING

print(AsyncResult(str(task_id)).state)
# SUCCESS

如何在Celery link_error回调中获得“完整”异步结果

3 个答案: