有效使用Multi-table继承（一对一关系）

Question

我需要几个模型，它们以一对一的关系继承自基础 class。与 Django 示例保持一致：

from django.db import models

class Place(models.Model):
    name = models.CharField(max_length=50)

class Restaurant(Place):
    serves_hot_dogs = models.BooleanField(default=False)
    serves_pizza = models.BooleanField(default=False)

class Garage(Place):
    car_brands_serviced = Models.ManyToManyField(Brands)

class Boutique(Place):
    for = Models.ChoiceField(choices=( ("m", "men"), ("w", "women"), ("k","kids"))

# etc

现在，当我在模板（或视图函数）中迭代各种类型的 Places 时，如何有效区分它们？现在，我只看到这个解决方案（如果我想遍历地点，而不是单独的子模型）：

for place in Place.objects.all():
    try:
        r = place.restaurant
        # Do restaurant stuff here
    except ObjectDoesNotExist:
        try:
            g = place.garage
            # Do garage stuff here
        except ObjectDoesNotExist:
            try:
                b = place.boutique
                # Do boutique stuff here
            except ObjectDoesNotExist:
                # Place is not specified

甚至不确定如何将其转化为模板，但这段代码似乎非常错误且效率低下。

作为一种逃避，我想你可以在 Place 中创建一个选择字段来跟踪哪个子模型是相关的，但这等同于危险的非规范化。

我是不是想多了？你是怎么做到的？

Answer 1

可以像这样简单吗：

models.py:

from django.db import models

class Place(models.Model):
    name = models.CharField(max_length=50)

class Restaurant(Place):
    serves_hot_dogs = models.BooleanField(default=False)
    serves_pizza = models.BooleanField(default=False)
    is_restaurant = True

class Garage(Place):
    car_brands_serviced = Models.ManyToManyField(Brands)
    is_garage = True

模板可以像这样工作 – template.html:

{% for place in places %}
 {% if place.is_restaurant %}
  <!-- Restaurant Stuff -->
 {% elif place.is_garage %}
  <!-- Garage Stuff -->
 {% endif %}
{% endfor %}

有效使用Multi-table继承（一对一关系）

Effectively use Multi-table inheritance (one-to-one relationships)

django

inheritance

model

multi-table-inheritance