为什么我不能通过应用程序路径启动芹菜

Question

我有一个 tasks.py 文件，我想将其提供给 Celery，以便它可以运行在后台运行。当我 cd 进入目录时，一切正常

celery -A tasks worker --loglevel=info

但是我想在 bash 脚本中运行这个，所以我必须指定应用程序的完整路径。但它似乎不起作用，即使路径是正确的

(tf_tasks) arthur@debian:~$ ls /data/tera_1/partage/Biomass_ML/
LICENSE  model.hdf5  __pycache__  server.py  tasks.py
(tf_tasks) arthur@debian:~$ celery -A "/data/tera_1/partage/Biomass_ML/tasks" worker --loglevel=info
Traceback (most recent call last):
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/bin/celery", line 10, in <module>
    sys.exit(main())
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/__main__.py", line 16, in main
    _main()
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/bin/celery.py", line 322, in main
    cmd.execute_from_commandline(argv)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/bin/celery.py", line 496, in execute_from_commandline
    super(CeleryCommand, self).execute_from_commandline(argv)))
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/bin/base.py", line 273, in execute_from_commandline
    argv = self.setup_app_from_commandline(argv)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/bin/base.py", line 479, in setup_app_from_commandline
    self.app = self.find_app(app)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/bin/base.py", line 501, in find_app
    return find_app(app, symbol_by_name=self.symbol_by_name)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/app/utils.py", line 359, in find_app
    sym = symbol_by_name(app, imp=imp)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/bin/base.py", line 504, in symbol_by_name
    return imports.symbol_by_name(name, imp=imp)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/kombu/utils/imports.py", line 56, in symbol_by_name
    module = imp(module_name, package=package, **kwargs)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/site-packages/celery/utils/imports.py", line 104, in import_from_cwd
    return imp(module, package=package)
  File "/data/tera_1/partage/anaconda/envs/tf_tasks/lib/python3.6/importlib/__init__.py", line 126, in import_module
    return _bootstrap._gcd_import(name[level:], package, level)
  File "<frozen importlib._bootstrap>", line 994, in _gcd_import
  File "<frozen importlib._bootstrap>", line 971, in _find_and_load
  File "<frozen importlib._bootstrap>", line 953, in _find_and_load_unlocked
ModuleNotFoundError: No module named '/data/tera_1/partage/Biomass_ML/tasks'

我尝试使用双引号、单引号、指定 --app= 而不是 -A、指定文件扩展名，但没有任何效果。

Answer 1

当使用 -A(--app) 启动 celery worker 时，我们应该将 python 路径传递给应用程序实例。例如

从/data/tera_1/partage/Biomass_ML/目录，我们可以运行

celery -A tasks worker --loglevel=info

当从 shell 脚本运行ning 时，您可以 cd 进入目录并运行它

cd /data/tera_1/partage/Biomass_ML/
celery -A tasks worker --loglevel=info

如果您必须从其他目录运行，请将其添加到您的 python 路径

export PYTHONPATH="/data/tera_1/partage/Biomass_ML/:$PYTHONPATH"
celery -A tasks worker --loglevel=info

或者，如果您的整个路径是包含模块的有效 python 路径，那么我们可以运行它

celery -A data.tera_1.partage.Biomass_ML.tasks worker --loglevel=info

为什么我不能通过应用程序路径启动芹菜

Why I can't start celery with a path to an app

path

celery