python - 如何获取 Django 项目的所有应用程序中某个类的所有模型

Question

我有一个模型类AppModelActions，它继承自CustomModel，而CustomModel又继承自django.db.models，并且我在所有应用程序中使用它项目的情况。

我想从其中一个应用程序的命令文件中获取所有应用程序中 AppModelActions 的所有实例。我无法显式导入每个命令，因为我希望此命令动态工作。我尝试以编程方式导入实例，但它不起作用(我收到此错误:`KeyError:'AppModelActions')。

这是我的代码:

from django.core.management.base import BaseCommand
from django.db import IntegrityError
from django.contrib.contenttypes.models import ContentType
from django.utils.translation import ugettext_lazy as _
from django.conf import settings
from GeneralApp import models

class Command(BaseCommand):
    help = _("""Run this commando to populate an empty database with initial data required for Attractora to work.
                For the momment, this models are considered on 'populate_db' command:
                    - InventoriManagerApp.InventoryManagerAppActions """)

    def populate_db(self):
        applications = settings.INSTALLED_APPS

        for application in applications:
            try:
                import applications.models
            except:
                pass

        #from InventoryManagerApp.models import InventoryManagerAppActions

        models_to_populate = (vars()['AppModelActions'])
        #models_to_populate = (InventoryManagerAppActions,)

        for model_to_populate in models_to_populate:
            self.populate_model('InventoryManagerApp', model_to_populate)

    def populate_model(self, app_label, model_to_populate):
        model_to_populate_instance = model_to_populate()
        content_type = ContentType.objects.get(app_label=app_label, model=model_to_populate_instance.name)

        if hasattr(model_to_populate_instance, 'populate_data'):
            populate_data = model_to_populate.populate_data
            for record in populate_data:
                for key, value in record.items():
                    setattr(model_to_populate_instance, key, value)
                try:
                    model_to_populate_instance.save()
                except IntegrityError:
                    print("{} already exists.".format(model_to_populate_instance))
                # else:
                #     new_permission = Permission.objects.create(code_name=)

    def handle(self, *args, **kwargs):
        self.populate_db()

Answer 1

您可以通过迭代更新一组子类来获取所有子类:

from django.core.management.base import BaseCommand

class Command(BaseCommand):

    # ...

    def populate_db(self):
        from some_app.models import AppModelActions

        children = set()
        gen = [AppModelActions]
        while gen:
            children.update(gen)
            gen = [sc for c in gen for sc in c.__subclasses__()]

        # do something with children
        # ...

此后，children是一个包含 AppModelActions 子类的所有模型的集合(包括该型号)。如果您不想包含 AppModelActions ，交换 while 中的两行循环。

既然你使用命令，Django会首先加载所有models.py来自已安装应用程序的文件，因此子类在 handle(..) 之前注册函数被执行，所以 children set 将包含已安装应用程序中的所有子模型。

请注意，子模型可以是抽象的，因此您可能需要执行额外的过滤，以便仅使用非抽象模型。例如，对其进行后处理:

non_abstract_children = [c for c in children if not c._meta.abstract]