我有两个型号List和Card。我正在尝试将两者结合起来并做出一个JSON响应
我的列表JSON响应
[
{
"id": 1,
"name": "List of things to do"
},
{
"id": 2,
"name": "one more"
}
]
我的卡JSON响应
[
{
"id": 1,
"title": "My first scrum card",
"description": "list things todo here",
"story_points": null,
"business_value": null,
"list": 1
},
{
"id": 2,
"title": "File my taxes",
"description": "fill it before 1st of nov",
"story_points": null,
"business_value": null,
"list": 1
},
]
我的serializers.py文件
from rest_framework import serializers
from .models import List, Card
class CardSerializer(serializers.ModelSerializer):
class Meta:
model = Card
fields = '__all__'
class ListSerializer(serializers.ModelSerializer):
cards = CardSerializer(read_only=True, many=True)
class Meta:
model = List
fields ='__all__'
我的api.py文件
from rest_framework.viewsets import ModelViewSet
from .models import List, Card
from .serializers import ListSerializer, CardSerializer
class ListViewSet(ModelViewSet):
queryset = List.objects.all()
serializer_class = ListSerializer
class CardViewSet(ModelViewSet):
queryset = Card.objects.all()
serializer_class = CardSerializer
我的模型。py
from django.db import models
class List(models.Model):
name = models.CharField(max_length=50)
def __str__(self):
return "List : {}".format(self.name)
class Card(models.Model): # to create card table
title = models.CharField(max_length=100)
description = models.TextField(blank=True)
list = models.ForeignKey(List, related_name = "card"
,on_delete=models.PROTECT) # creating a foriegn key for storing list
story_points = models.IntegerField(null=True, blank = True)
business_value = models.IntegerField(null=True, blank = True)
def __str__(self):
return "Card : {}".format(self.title)
如何在“我的列表” JSON响应中实现以下内容?
[
{
"id": 1,
"cards":[
{
"id": 1,
"title": "My first scrum card",
"description": "list things todo here",
"story_points": null,
"business_value": null,
"list": 1
}],
"name": "List of things to do"
},
{
"id": 2,
"cards":[
{
"id": 2,
"title": "File my taxes",
"description": "fill it before 1st of nov",
"story_points": null,
"business_value": null,
"list": 1
}],
"name": "one more"
},
]
我尝试这样做,但是无法实现。所以我在这里张贴了
非常感谢。
答案 0 :(得分:2)
将列表related_name
更改为cards
,ListViewSet
会做您想要的事
class Card(models.Model): # to create card table
title = models.CharField(max_length=100)
description = models.TextField(blank=True)
list = models.ForeignKey(List, related_name = "cards"
,on_delete=models.PROTECT)
...
答案 1 :(得分:0)
您还可以使用SerializerMethodeField https://www.django-rest-framework.org/api-guide/fields/#serializermethodfield。
class ListSerializer(serializers.ModelSerializer):
cards = serializers.SerializerMethodField()
class Meta:
model = List
fields = '__all__'
def get_cards(self, obj):
data = CardSerializer(obj.card.all(), many=True).data
return data