提问人:Azima 提问时间:4/15/2021 更新时间:4/16/2021 访问量:17604
python celery 无效值 -A 无法加载应用程序
python celery invalid value for -A unable to load application
问:
我有一个以下项目目录:
azima:
__init.py
main.py
tasks.py
task.py
from .main import app
@app.task
def add(x, y):
return x + y
@app.task
def mul(x, y):
return x * y
@app.task
def xsum(numbers):
return sum(numbers)
main.py
from celery import Celery
app = Celery('azima', backend='redis://localhost:6379/0', broker='redis://localhost:6379/0', include=['azima.tasks'])
# Optional configuration, see the application user guide.
app.conf.update(
result_expires=3600,
)
if __name__ == '__main__':
app.start()
当我运行命令初始化芹菜工人时,出现以下错误:
(azima_venv) brightseid@darkseid:~$ celery -A azima worker -l INFO
Usage: celery [OPTIONS] COMMAND [ARGS]...
Error: Invalid value for '-A' / '--app':
Unable to load celery application.
Module 'azima' has no attribute 'celery'
但是当我重命名为之前的样子时,没有问题。main.py
celery.py
我在这里错过了什么?
答:
8赞
Patricio
4/16/2021
#1
有两种方法
- 将你的导入到
app
azima/__init__.py
from azima.main import app
celery = app # you can omit this line
您可以省略最后一行,celery 将从导入中识别出芹菜应用程序。然后你打电话给celery -A azima worker -l INFO
- 称您为应用程序
celery -A azima.main worker -l INFO
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