使用 get_context_data() 方法检索 DetailView Django 中的依赖对象列表

Question

我正在 Django 3 中创建一个简单的汽车经销商应用程序，当转到某个类别的详细信息页面时，它应该列出属于该类别的所有汽车，因此在我的情况下，汽车对象是它们各自类别的依赖对象。我尝试使用 get_context_data() 方法来实现这一点，并在 DetailView 中引用两个各自的模型，这是我的代码

models.py 类别

from django.db import models

from django.urls import reverse

# Create your models here.


class Category(models.Model):
    name = models.CharField(null=False, blank=False, max_length=20)
    description = models.TextField(null=False, blank=False, max_length=120)
    image = models.ImageField(upload_to='images/')

    def __str__(self):
        return self.name

    def get_absolute_url(self):
        return reverse('category_detail', args=[str(self.id)])

models.py 汽车

from django.db import models

from django.urls import reverse

from categories.models import Category

# Create your models here.


class Car(models.Model):
    image = models.ImageField(upload_to='images/cars/')
    make = models.CharField(null=False, blank=False, max_length=30)
    model = models.CharField(null=False, blank=False, max_length=30)
    year = models.IntegerField(null=False, blank=False)
    transmission = models.CharField(null=False, blank=False, max_length=30)
    category = models.ForeignKey(Category, on_delete=models.CASCADE)

    def __str__(self):
        return self.model

    def get_absolute_url(self):
        return reverse('car_detail', args=[str(self.id)])

查看类别

from django.shortcuts import render

from django.views.generic import ListView, DetailView

from .models import Category
from cars.models import Car
# Create your views here.


class CategoryList(ListView):
    model = Category
    template_name = 'category_list.html'


class CategoryDetailView(DetailView):
    model = Category
    template_name = 'category_detail.html'

    def get_context_data(self, *args, **kwargs):
        context = super(CategoryDetailView, self).get_context_data(
            *args, **kwargs)
        context['category_cars'] = Car.objects.filter(
            category=self.request.name)
        return context

网址类别

from django.urls import path

from .views import CategoryList, CategoryDetailView

urlpatterns = [
    path('categories/', CategoryList.as_view(), name='categories'),
    path('categories/<int:pk>/', CategoryDetailView.as_view(), name='category_detail')
]

所以使用我上面的实现会导致错误

有什么建议可以实现我想要的功能吗？

Answer 1

参数通过kwargs传递，你的参数名称是pk而不是name

def get_context_data(self, *args, **kwargs):
    context = super(CategoryDetailView, self).get_context_data(
        *args, **kwargs)
    context['category_cars'] = Car.objects.filter(
        category=context['object']
    )
    return context

但您实际上不需要这样做，因为 Category 对象可以访问它们的 related Car 对象 (car_set)：

{% for car in object.car_set.all %}
   {{car.make}}
{% endfor %}