python - Django REST 组外键模型

Question

我有如下所示的模型。

菜单模型

class Menu(models.Model):

    name = models.CharField(max_length=40, unique=True, verbose_name='menu name')

商品型号

class Item(models.Model):

    shop = models.ForeignKey(Shop)
    menu = models.ForeignKey(Menu)

    name = models.CharField(max_length=500)
    price = models.IntegerField(default=0)

我想获取商店 ID 的菜单。

Item.objects.filter(shop_id = 1)

如何按商店 ID 1 的菜单名称对结果进行分组？

示例。

{
 menu: menuname1
 items: [ {item1}, {item2}]
},
{
 menu: menuname2
 items: [ {item1}, {item2}]
}

谢谢..

Answer 1

您可以使用嵌套序列化器获得所需的输出，如下所示:

型号

class Item(models.Model):
    class Meta:
        unique_together = ('shop', 'menu',)

    shop = models.ForeignKey(Shop)
    menu = models.ForeignKey(Menu, related_name='items')
    name = models.CharField(max_length=500)
    price = models.IntegerField(default=0)

序列化器

class ItemSerializer(serializers.ModelSerializer):
    class Meta:
        model = Item
        fields = '__all__'

class MenuSerialzer(serializers.ModelSerializer):
    items = ItemSerializer(many=True, read_only=True)
    class Meta:
        model = Menu
        fields = ('name','items')

观看次数

class MenuList(generics.ListAPIView):
    serializer_class = MenuSerializer

    def get_queryset(self):
        return Menu.objects.filter(item__shop__id = 1)

我确信你不希望每次都得到 shop_id = 1所以这样做:

class MenuList(generics.ListAPIView):
    def get_queryset(self):
        shop_id = int(self.kwargs['shop_id'])
        return Menu.objects.filter(item__shop__id = 1)

url.py

urlpatterns = [
   url('^menus/(?P<shop_id>[0-9]+)/$', MenuList.as_view()),
]