Python Django：使用模型继承时如何检查对象类别？

这是我的

models.py

应用程序的理念是让不同的对象（餐厅、服务等）具有不同的功能（您可以检查每个模型）。但是当我想获取所有对象时，我不知道每个对象的类别是什么

models.py：

from django.db import models
from users.models import ProfileUser
from django.utils import timezone

# Create your models here.

class City(models.Model):
    name = models.CharField(max_length=20)

    def __str__(self):
        return f"{self.name}"

class Object(models.Model):
    author = models.ForeignKey(ProfileUser, on_delete=models.CASCADE)
    title = models.CharField(max_length=300)
    city = models.ForeignKey(City, on_delete=models.CASCADE)
    address = models.CharField(max_length=300)
    phone = models.CharField(max_length=20, default='')
    email = models.CharField(max_length=100, default='')
    site = models.CharField(max_length=100, default='')
    facebook = models.CharField(max_length=100, default='')
    instagram = models.CharField(max_length=100, default='')
    content = models.TextField()
    rating = models.IntegerField(default=10)
    created_date = models.DateTimeField(default=timezone.now)
    approved_object = models.BooleanField(default=False)
    admin_seen = models.BooleanField(default=False)

    def __str__(self):
        return f"{self.title}"


class Restaurant(Object):
    seats = models.IntegerField()
    bulgarian_kitchen = models.BooleanField(default=False)
    italian_kitchen = models.BooleanField(default=False)
    french_kitchen = models.BooleanField(default=False)
    is_garden = models.BooleanField(default=False)
    is_playground = models.BooleanField(default=False)


class SportFitness(Object):
    is_fitness_trainer = models.BooleanField(default=False)


class CarService(Object):
    is_parts_clients = models.BooleanField(default=False)


class BeautySalon(Object):
    is_hair_salon = models.BooleanField(default=False)
    is_laser_epilation = models.BooleanField(default=False)


class FastFood(Object):
    is_pizza = models.BooleanField(default=False)
    is_duner = models.BooleanField(default=False)
    is_seats = models.BooleanField(default=False)


class CarWash(Object):
    is_external_cleaning = models.BooleanField(default=False)
    is_internal_cleaning = models.BooleanField(default=False)
    is_engine_cleaning = models.BooleanField(default=False)


class Fun(Object):
    is_working_weekend = models.BooleanField(default=False)
    is_kids_suitable = models.BooleanField(default=False)


class Other(Object):
    is_working_weekend = models.BooleanField(default=False)


class Comment(models.Model):
    object = models.ForeignKey(Object, on_delete=models.CASCADE, related_name='comments')
    author = models.ForeignKey(ProfileUser, on_delete=models.CASCADE)
    content = models.TextField()
    rating = models.TextField()
    created_date = models.DateTimeField(default=timezone.now)

    def __str__(self):
        return f"{self.content}"

---

无法使用对象模型获取类别。 对象类是为特定目的创建的： 它无需声明类的某些特征

在这里，您可以使用select_related或PERFICT_related获得所需的输出

或

为了满足您的需求，您可以使用django Polymopic了解更多详细信息，您可以查看文档。

我就是这样做的：

基于我实现的@wfehr commond和@Lokesh Pathak answer，还为每个模型添加了两行（列）（目前仅适用于餐厅），因此当foreach覆盖所有对象时，我可以打印类别的确切名称，并添加了其他字段

category_en_name = models.CharField(max_length=100, default='restaurants')
category_bg_name = models.CharField(max_length=100, default='Ресторант')

完整的

型号.py

：

from django.db import models
from users.models import ProfileUser
from django.utils import timezone
from polymorphic.models import PolymorphicModel

# Create your models here.

class City(models.Model):
    name = models.CharField(max_length=20)

    def __str__(self):
        return f"{self.name}"

class Object(PolymorphicModel):
    author = models.ForeignKey(ProfileUser, on_delete=models.CASCADE)
    title = models.CharField(max_length=300)
    city = models.ForeignKey(City, on_delete=models.CASCADE)
    address = models.CharField(max_length=300)
    phone = models.CharField(max_length=20, default='')
    email = models.CharField(max_length=100, default='')
    site = models.CharField(max_length=100, default='')
    facebook = models.CharField(max_length=100, default='')
    instagram = models.CharField(max_length=100, default='')
    content = models.TextField()
    rating = models.IntegerField(default=10)
    created_date = models.DateTimeField(default=timezone.now)
    approved_object = models.BooleanField(default=False)
    admin_seen = models.BooleanField(default=False)

    def __str__(self):
        return f"{self.title}"


class Restaurant(Object):
    seats = models.IntegerField()
    bulgarian_kitchen = models.BooleanField(default=False)
    italian_kitchen = models.BooleanField(default=False)
    french_kitchen = models.BooleanField(default=False)
    category_en_name = models.CharField(max_length=100, default='restaurants')
    category_bg_name = models.CharField(max_length=100, default='Ресторант')
    is_garden = models.BooleanField(default=False)
    is_playground = models.BooleanField(default=False)


class SportFitness(Object):
    is_fitness_trainer = models.BooleanField(default=False)

因此，我只能在views.py中选择餐厅：

objects = Object.objects.instance_of(Restaurant)

在模板中，我可以显示每个餐厅的类别名称：

{% for object in objects %}
{{ object.category_bg_name }}
{% endfor %}

---

django多态性可能是您想要查看的东西。->对于

select_related

I应该向对象类添加额外的行（列），对于category right？@MorganFreeFarm No，如果使用django多态，您可以添加额外的行（列）。但与选择无关。例如：-Objects.Objects.select_related（“author”）在一个查询中提供“author”所在的所有表的数据。@MorganFreeFarm如果我的答案有帮助，请投赞成票