python - 如何在 django 命令中实现应用程序命名空间？

Question

我想在像这样的命令中实现 django 应用程序命名空间

python manage.py appname:command --parameters

和

python manage.py appname:command:subcommand --parameters

我怎么能做到这一点？

Answer 1

Django 管理命令基于 Python 标准库中的 argparse(自 Django 1.10 [1] 起)。一种可能的解决方案是使用 subparser。 [2]
首先创建一个文件作为标准管理命令，例如:yourapp/management/commands/base.py .

from __future__ import annotations

from typing import Any, Callable, Dict

from django.core.management.base import BaseCommand


def create_resource(**options):
    print("Creating resource %s" % options["name_of_resource_to_create"])


def delete_resource(**options):
    print("Deleting resource %s" % options["name_of_resource_to_delete"])


SUBCOMMANDS_MAPPING: Dict[str, Callable[[Dict[str, Any]], None]] = {
    "create_resource": create_resource,
    "delete_resource": delete_resource,
}


class Command(BaseCommand):
    def add_arguments(self, parser):
        subparsers = parser.add_subparsers(
            dest="subcommand",
            title="positional arguments:",
            description="Required positional arguments to specify subcommand to execute.",
            required=True,
        )

        # Parser for create_resource:
        parser_serve = subparsers.add_parser(name="create_resource")
        parser_serve.add_argument("name_of_resource_to_create", type=str)

        # Parser for delete_resource:
        parser_serve = subparsers.add_parser(name="delete_resource")
        parser_serve.add_argument("name_of_resource_to_delete", type=str)

    def handle(self, *args, **options):
        print('Base command')
        SUBCOMMANDS_MAPPING[options.pop("subcommand")](**options)

我们添加了一个名为 subcommand 的子解析器.每个子命令都有自己单独的必需/可选参数。子命令名称将包含在 subcommand 中 key 。根据这个值，我们决定运行哪个函数。
这是帮助输出的样子:

$ python manage.py base --help

usage: manage.py base [-h] [--version] [-v {0,1,2,3}] [--settings SETTINGS] [--pythonpath PYTHONPATH] [--traceback] [--no-color] [--force-color] [--skip-checks] {create_resource,delete_resource} ...

optional arguments:
  -h, --help            show this help message and exit
  --version             show program's version number and exit
  -v {0,1,2,3}, --verbosity {0,1,2,3}
                        Verbosity level; 0=minimal output, 1=normal output, 2=verbose output, 3=very verbose output
  --settings SETTINGS   The Python path to a settings module, e.g. "myproject.settings.main". If this isn't provided, the DJANGO_SETTINGS_MODULE environment variable will be used.
  --pythonpath PYTHONPATH
                        A directory to add to the Python path, e.g. "/home/djangoprojects/myproject".
  --traceback           Raise on CommandError exceptions
  --no-color            Don't colorize the command output.
  --force-color         Force colorization of the command output.
  --skip-checks         Skip system checks.

positional arguments::
  Required positional arguments to specify subcommand to execute.

  {create_resource,delete_resource}


$ python manage.py create_resource --help

usage: manage.py base create_resource [-h] name_of_resource_to_create

positional arguments:
  name_of_resource_to_create

optional arguments:
  -h, --help            show this help message and exit

我在 Django 3.1 和 Python 3.9 中测试了这个解决方案。我的灵感来源是这个答案: https://stackoverflow.com/a/37414551/752142 .

[1] https://docs.djangoproject.com/en/dev/internals/deprecation/#deprecation-removed-in-1-10

[2] https://docs.python.org/3/library/argparse.html#sub-commands