属性错误：“Flask”对象没有属性“user_options”

Question

我正在尝试从以下文档中设置这个基本示例：

http://flask.pocoo.org/docs/patterns/celery/

但到目前为止我一直收到以下错误：

AttributeError：“Flask”对象没有属性“user_options”

我正在使用芹菜3.1.15。

from celery import Celery

def make_celery(app):
    celery = Celery(app.import_name, broker=app.config['CELERY_BROKER_URL'])
    celery.conf.update(app.config)
    TaskBase = celery.Task
    class ContextTask(TaskBase):
        abstract = True
        def __call__(self, *args, **kwargs):
            with app.app_context():
                return TaskBase.__call__(self, *args, **kwargs)
    celery.Task = ContextTask
    return celery

示例：

from flask import Flask

app = Flask(__name__)
app.config.update(
    CELERY_BROKER_URL='redis://localhost:6379',
    CELERY_RESULT_BACKEND='redis://localhost:6379'
)
celery = make_celery(app)


@celery.task()
def add_together(a, b):
    return a + b

回溯错误：

Traceback (most recent call last):
  File "/usr/local/bin/celery", line 11, in <module>
    sys.exit(main())
  File "/usr/local/lib/python2.7/dist-packages/celery/__main__.py", line 30, in main
    main()
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/celery.py", line 81, in main
    cmd.execute_from_commandline(argv)
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/celery.py", line 769, in execute_from_commandline
    super(CeleryCommand, self).execute_from_commandline(argv)))
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 305, in execute_from_commandline
    argv = self.setup_app_from_commandline(argv)
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 473, in setup_app_from_commandline
    user_preload = tuple(self.app.user_options['preload'] or ())
AttributeError: 'Flask' object has no attribute 'user_options'

Answer 1

基于 Flask Celery 的后台任务页面 (http://flask.pocoo.org/docs/patterns/celery/) 建议这样启动 celery:

celery -A your_application worker

your_application 字符串必须指向创建 celery 对象的应用程序的包或模块。

假设代码位于 application.py 中，显式指向 celery 对象（不仅仅是模块名称）可以避免错误：

celery -A application.celery worker

Answer 2

这对我有用：

celery -A my_app_module_name.celery worker

Answer 3

重命名应用程序flask_app 会起作用的

Answer 4

Celery 试图运行错误

app

它是 Flask 实例
solotion1：重命名

app -> flask_app

或其他名称
解决方案2：指定

celery -A my_app_module_name.celery worker

（已添加

.celery

）

Answer 5

像这样：

celery -A your_application worker

你的应用程序所在的位置：

your_application = Flask(\__name\__)

python文件名：your_application.py，它会工作

顺便说一下，Windows 不支持 celery v4