无法使用 Celery + Django 在 tasks.py 中导入模型

Can't import models in tasks.py with Celery + Django

我想创建一个后台任务来更新特定日期的记录。我在 RabbitMQ 中使用 Django 和 Celery。

我已经设法在使用此虚拟任务函数保存模型时调用任务:

tasks.py

from __future__ import absolute_import
from celery import Celery
from celery.utils.log import get_task_logger

logger = get_task_logger(__name__)

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


@app.task(name='news.tasks.update_news_status')
def update_news_status(news_id):
    # (I pass the news id and return it, nothing complicated about it)
    return news_id

此任务是从我的 models.py

中的 save() 方法调用的
from django.db import models
from celery import current_app


class News(models.model):
    (...)

    def save(self, *args, **kwargs):

        current_app.send_task('news.tasks.update_news_status', args=(self.id,))

        super(News, self).save(*args, **kwargs)

事情是 我想在 tasks.py 中导入我的新闻模型,但如果我想这样:

from .models import News

我收到这个错误:

django.core.exceptions.ImproperlyConfigured: Requested setting DEFAULT_INDEX_TABLESPACE, but settings are not configured. You must either define the environment variable DJANGO_SETTINGS_MODULE or call settings.configure() before accessing settings.

这就是 mi celery.py 的样子

from __future__ import absolute_import, unicode_literals
from celery import Celery
import os
# set the default Django settings module for the 'celery' program.
os.environ.setdefault('DJANGO_SETTINGS_MODULE', 'myapp.settings')
app = Celery('myapp')

# Using a string here means the worker doesn't have to serialize
# the configuration object to child processes.
# - namespace='CELERY' means all celery-related configuration keys
#   should have a `CELERY_` prefix.
app.config_from_object('django.conf:settings', namespace='CELERY')

# Load task modules from all registered Django app configs.
app.autodiscover_tasks()

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

我已经试过了:

  1. can't import django model into celery task
  2. 我试过在任务方法中进行导入
  3. 这个我也试过了
  4. 我也试过创建 utils.py 并导入它,但没成功。

和运行进入不同的错误,但最后我无法在tasks.py

中导入任何模块

我的配置可能有问题,但我看不到错误,我按照 The Celery Docs: First steps with Django

中的步骤操作

此外,我的项目结构如下所示:

├── myapp
│   ├── __init__.py
├── ├── celery.py
│   ├── settings.py
│   ├── urls.py
│   └── wsgi.py
├── news
│   ├── __init__.py
│   ├── admin.py
│   ├── apps.py
│   ├── tasks.py
│   ├── urls.py
│   ├── models.py
│   ├── views.py
├── manage.py

我正在从 myapp 目录执行 worker,如下所示:

celery -A news.tasks worker --loglevel=info

我在这里错过了什么?预先感谢您的帮助!

拉姆达:settings.INSTALLED_APPS

编辑

进行评论中建议的更改后: 将此添加到 celery.py app.autodiscover_tasks(lambda: settings.INSTALLED_APPS)

并导入内部方法:tasks.py

from __future__ import absolute_import
from celery import Celery
from celery.utils.log import get_task_logger

logger = get_task_logger(__name__)

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


@app.task(name='news.tasks.update_news_status')
def update_news_status(news_id):
    from .models import News
    return news_id

我收到以下错误:

[2018-07-20 12:24:29,337: ERROR/ForkPoolWorker-1] Task news.tasks.update_news_status[87f9ec92-c260-4ee9-a3bc-5f684c819f79] raised unexpected: ValueError('Attempted relative import in non-package',)
Traceback (most recent call last):
  File "/Users/carla/Develop/App/backend/myapp-venv/lib/python2.7/site-packages/celery/app/trace.py", line 382, in trace_task
    R = retval = fun(*args, **kwargs)
  File "/Users/carla/Develop/App/backend/myapp-venv/lib/python2.7/site-packages/celery/app/trace.py", line 641, in __protected_call__
    return self.run(*args, **kwargs)
  File "/Users/carla/Develop/App/backend/news/tasks.py", line 12, in update_news_status
    from .models import News
ValueError: Attempted relative import in non-package

尝试这样的事情。它在 3.1 celery 中工作,导入应该发生在保存方法内部和 super()

之后
from django.db import models



class News(models.model):
    (...)

    def save(self, *args, **kwargs):
        (...)
        super(News, self).save(*args, **kwargs)
        from task import update_news_status
        update_news_status.apply_async((self.id,)) #apply_async or delay

好吧,对于任何为此苦苦挣扎的人...原来我的 celery.py 没有从设置中读取环境变量。

经过一周的大量研究,我意识到 Celery 不是 Django 的一个进程,而是一个 运行 在它之外的进程 (duh),所以当我尝试加载它们已加载的设置,但后来我无法访问我在 .env 中定义的环境变量(我使用 dotenv 库)。 Celery 试图在我的 .bash_profile 中查找环境变量(当然)

所以最后我的解决方案是在定义我的 celery.py 的同一目录中创建一个辅助模块,称为 load_env.py 和以下

from os.path import dirname, join
import dotenv


def load_env():
    "Get the path to the .env file and load it."
    project_dir = dirname(dirname(__file__))
    dotenv.read_dotenv(join(project_dir, '.env'))

然后在我的 celery.py 上(注意最后导入和第一条指令)

from __future__ import absolute_import, unicode_literals
from celery import Celery
from django.conf import settings
import os
from .load_env import load_env

load_env()

# set the default Django settings module for the 'celery' program.
os.environ.setdefault("DJANGO_SETTINGS_MODULE", "myapp.settings")

app = Celery('myapp')

# Using a string here means the worker doesn't have to serialize
# the configuration object to child processes.
# - namespace='CELERY' means all celery-related configuration keys
#   should have a `CELERY_` prefix.

app.config_from_object('myapp.settings', namespace='CELERY')

# Load task modules from all registered Django app configs.
app.autodiscover_tasks(lambda: settings.INSTALLED_APPS)

在调用 load_env() env 变量后加载并且 celery worker 可以访问它们。通过这样做 我现在可以从我的 tasks.py 访问其他模块,这是我的主要问题。

感谢 this guys (Caktus Consulting Group) and their django-project-template,因为如果不是他们,我将找不到答案。谢谢。

这是我会做的(Django 1.11 和 celery 4.2),你的 celery 配置有问题,你尝试重新声明 Celery 实例:

tasks.py

from myapp.celery import app # would contain what you need :)
from celery.utils.log import get_task_logger

logger = get_task_logger(__name__)

@app.task(name='news.tasks.update_news_status')
def update_news_status(news_id):
    # (I pass the news id and return it, nothing complicated about it)
    return news_id

celery.py

from __future__ import absolute_import, unicode_literals
from celery import Celery
from django.conf import settings
import os

os.environ.setdefault("DJANGO_SETTINGS_MODULE", "myapp.settings")
app = Celery('myapp', backend='rpc://', broker=BROKER_URL) # your config here
app.config_from_object('django.myapp:settings', namespace='CELERY') # change here
app.autodiscover_tasks()

models.py

from django.db import models

class News(models.model):
    (...)
    def save(self, *args, **kwargs):
        super(News, self).save(*args, **kwargs)
        from news.tasks import update_news_status
        update_news_status.delay(self.id) # change here

并使用 celery -A myapp worker --loglevel=info 启动它,因为您的应用程序是在 myapp.celery 中定义的,所以 -A 参数需要是声明 conf 的应用程序