Python Django REST序列化输出-按外键分组
我有下面这样的模型 餐厅模式Python Django REST序列化输出-按外键分组,python,django,django-models,django-rest-framework,Python,Django,Django Models,Django Rest Framework,我有下面这样的模型 餐厅模式 class Restaurant(models.Model): name = models.CharField(max_length=40, verbose_name='Name') 菜单模型 class Menu(models.Model): name = models.CharField(max_length=40, unique=True, verbose_name='menu name') 项目模型 class Item(models
class Restaurant(models.Model):
name = models.CharField(max_length=40, verbose_name='Name')
菜单模型
class Menu(models.Model):
name = models.CharField(max_length=40, unique=True, verbose_name='menu name')
项目模型
class Item(models.Model):
restaurant = models.ForeignKey(Restaurant)
menu = models.ForeignKey(Menu)
name = models.CharField(max_length=500)
price = models.IntegerField(default=0)
我想得到商店id的菜单
如何根据餐厅id的菜单对结果进行分组
呼叫获取/菜单/餐厅\u id
样品
{
name: menu name 1
items: [ {item1}, {item2}]
},
{
name: menu name 2
items: [ {item1}, {item2}]
}
谢谢..我唯一能找到的就是postgres特有的聚合功能 您可以这样使用它:
from django.contrib.postgres.aggregates import ArrayAgg
Item.objects.filter(restaurant_id=1).values('menu__name').annotate(items=ArrayAgg('name'))
# example output:
# [
# {
# 'menu__name': 'menu1',
# 'items': ['item1', 'item2']
# },
# {
# 'menu__name': 'menu2',
# 'items': ['item3', 'item4']
# },
# ]
此类qs执行下一个原始sql查询:
SELECT
"appname_menu"."name",
ARRAY_AGG("appname_item"."name") AS "items"
FROM "appname_item"
INNER JOIN "appname_menu" ON ("appname_item"."menu_id" = "appname_menu"."id")
WHERE "appname_item"."restaurant_id" = 1
GROUP BY "appname_menu"."name"
也许它能帮助你