在 Django 中通过子 class 验证父模型外键

Question

假设我的 Django 应用程序中有以下父模型：

class Location(models.Model):
    name = models.CharField(max_length=100)

class Exit(models.Model):
    location = models.ForeignKey(Location, on_delete=models.CASCADE, related_name="exits")
    closed = models.BooleanField()

以及两对对应的子模型：

class Submarine(Location):
    size = models.FloatField()


class Hatch(Exit):
    diameter = models.FloatField()

class House(Location):
    height = models.FloatField()


class Door(Exit):
    width = models.FloatField()
    height = models.FloatField()

在此设置中，House 可以将 Hatch 作为其 Exit 之一，并且 Submarine 可以将Door。有没有办法明确防止这种情况发生？理想情况下，我希望在尝试设置无效外键时抛出异常。

将 location 字段从 Exit 移动到 Hatch 和 Door 不是一个选项，因为我希望能够使用如下结构：

open_locations = Location.objects.filter(exits__closed=False)

并避免重复（即为 House 和 Submarine 编写单独的函数）。

也许 limit_choices_to 约束可能有用，但我没弄清楚如何在此处应用它。

Answer 1

您可以使用 CheckConstraint:

# models.py
from django.db import models

class Location(models.Model):
    name = models.CharField(max_length=100)

class Exit(models.Model):
    location = models.ForeignKey(Location, on_delete=models.CASCADE, related_name="exits")
    closed = models.BooleanField()

class Submarine(Location):
    size = models.FloatField()


class Hatch(Exit):
    diameter = models.FloatField()
    
    class Meta:
        constraints = [
            CheckConstraint(
                check=Q(location__submarine__is_null=False),
                name='hatches_must_be_submarine_exits'
            ),
        ]


class House(Location):
    height = models.FloatField()


class Door(Exit):
    width = models.FloatField()
    height = models.FloatField()

    class Meta:
        constraints = [
            CheckConstraint(
                check=Q(location__house__is_null=False),
                name='doors_must_be_house_exits'
            ),
        ]

编辑：一种更 DRY 的方法是在 python 级别而不是数据库级别进行验证。您可以像这样使用一个 clean-method：

# models.py
from django.db import models
from django.contrib.contenttypes.models import ContentType
from django.core.exceptions import ValidationError


class Location(models.Model):
    name = models.CharField(max_length=100)

class Exit(models.Model):
    location = models.ForeignKey(Location, on_delete=models.CASCADE, related_name="exits")
    location_type = ContentType.objects.get_for_model(Location)
    closed = models.BooleanField()

    def clean(self):
        if self.location is not None:
            actual_type = ContentType.objects.get_for_model(self.location.__class__)
            expected_type = self.__class__.location_type
            if (
                actual_type
                is not expected_type
            ):
                raise ValidationError(
                    message=f'location must be a {expected_type.name}, not a {actual_type.name}'
                )

    

class Submarine(Location):
    size = models.FloatField()


class Hatch(Exit):
    location_type = ContentType.objects.get_for_model(Submarine)
    diameter = models.FloatField()


class House(Location):
    height = models.FloatField()


class Door(Exit):
    location_type = ContentType.objects.get_for_model(House)
    width = models.FloatField()
    height = models.FloatField()

此外，您还可以限制显示的选项，例如，在实例化 Form 时：

from django import forms

from my_app import models as my_models

class ExitForm(forms.ModelForm):
    def __init__(self, *args, **kwargs):
        super().__init__(*args, **kwargs)
        loc_model = self._meta.model.location_type.model_class()
        self.fields['location'].choices = loc_model.objects.values_list('location__pk', 'name')


    class Meta:
        model = my_models.Exit
        fields = '__all__'

Answer 2

不幸的是，Django 在继承方面不是很好，因为每个 child 仍然需要自己的数据库 table。所以即使你可以完成这项工作，它看起来也不会很好，也不会帮助你走下去。最简单和最 Djangesque 的方式是

class Location(models.Model):
    name = models.CharField(max_length=100)

    class Meta:
        abstract = True

class Exit(models.Model):
    closed = models.BooleanField()

    class Meta:
        abstract = True

class Submarine(Location):
    size = models.FloatField()

class Hatch(Exit):
    diameter = models.FloatField()
    location = models.ForeignKey(Submarine, on_delete=models.CASCADE, related_name="exits")

class House(Location):
    height = models.FloatField()

class Door(Exit):
    width = models.FloatField()
    height = models.FloatField()
    location = models.ForeignKey(House, on_delete=models.CASCADE, related_name="exits")

我添加了 Meta 和 abstract = True 因为我的直觉是你不会想要任何普通的 Location 和 Exit objects数据库，但我可能错了； Meta.abstract 告诉 Django 你不需要 DB tables 用于抽象 parent 模型。重复的 Location 行很不幸，但如果有很多这样的模型，你最好使用工厂而不是继承。

看起来像这样：

class Exit(models.Model):
    closed = models.BooleanField()

    class Meta:
        abstract = True
    
    def location_field_factory(exit_type):
        assert isinstance(exit_type, Exit)
        return models.ForeignKey(exit_type, on_delete=models.CASCADE, related_name="exits")

class Barrel(Location):
    diameter = models.FloatField()
    height = models.FloatField()

class Lid(Exit):
    diameter = models.FloatField()
    location = Exit.location_field_factory(Barrel)

在 Django 中通过子 class 验证父模型外键

Validate parent model foreign key by child class in Django

python

django

validation

model