Django 如何访问模板中的不同模型？

我有三种型号

我是否需要更改模型中的连接，并使键字段不是id而是名称

class Category(models.Model):
    name = models.CharField(max_length=150, unique=True)
    description = models.CharField(max_length=250)

class Company(models.Model):
    name = models.CharField(max_length=150, unique=True)
    country = models.CharField(max_length=50)

class Motobike(models.Model):
    name = models.CharField(max_length=150)
    company = models.ForeignKey('Company', on_delete=models.CASCADE)
    category = models.ForeignKey('Category', on_delete=models.CASCADE)

和测试：

def test_category(setup):
    client = Client()
    category_id = Category.objects.get(name='Мотоциклы').id
    response = client.get(f'/categories/{category_id}/')
    assert response.status_code == 200
    response_data = json.loads(response.content.decode('utf-8'))
    assert len(response_data) == 2
    assert response_data[1]['name'] == 'Ninja Turbo'
    assert response_data[1]['vendor'] == 'Kawasaki'
    assert response_data[1]['category'] == 'Мотоциклы'
    assert response_data[1]['description'] == ''

    response = client.get(f'/categories/25/')
    assert response.status_code == 404

因此，我认为：

 class CategoryView(DetailView):
     model = Category
     template_name = 'bikes_site/categories_detail.html'

     def get_context_data(self, id, **kwargs):
        context = get_object_or_404(self.model, id)
        context['motobikes'] =   Motobike.objects.filter(category_id=id).all()
    return context

我得到一个错误：

get\u context\u data（）

缺少1个必需的位置参数：“id”

---

get\u context\u data

的函数签名错误，应该是错误的

def get_context_data(self, **kwargs):
    //todo

您的详细视图应该如下所示

 class CategoryView(DetailView):
     model = Category
     template_name = 'bikes_site/categories_detail.html'
     pk_url_kwarg = "id"

     def get_context_data(self, **kwargs):
       context = super().get_context_data(**kwargs) 
       category = self.get_object()
       context['motobikes'] =   Motobike.objects.filter(category_id=category.pk)
       return context

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多谢各位