我在Django REST框架中实现嵌套序列化器时遇到了困难。
我正在构建一个在线记分板,目前我有三个模型,我正试图将其序列化为一个响应。
模型.py
class Player(models.Model):
first_name = models.CharField(max_length=50)
last_name = models.CharField(max_length=50)
def __str__(self):
return self.first_name
class Event(models.Model):
user = models.ManyToManyField("Player")
name = models.CharField(max_length=50)
desc = models.CharField(max_length=225)
def __str__(self):
return self.name
class LeadBoard(models.Model):
"""model for LeadBoard"""
player = models.ForeignKey(Player, on_delete=models.CASCADE, related_name='leadboard_player', null=True, blank=True)
event = models.ForeignKey(Event, on_delete=models.CASCADE, related_name='leadboard_event', null=True, blank=True)
score = models.IntegerField()
def __str__(self):
return self.score
事件模型代表某种体育事件,每个事件可以有多个玩家(* 事件模型中的用户字段 *),排行榜模型存储每个事件中玩家的分数。
序列化程序.py
class PlayerSerializer(serializers.ModelSerializer):
class Meta:
model = Player
fields = ('id','first_name', 'last_name', 'bio')
class EventSerializer(serializers.ModelSerializer):
user = PlayerSerializer(read_only=True, many=True)
class Meta:
model = Event
fields = ('id', 'name', 'user', 'desc')
class LeadBoardSerializer(serializers.ModelSerializer):
event = EventSerializer(read_only=True)
class Meta:
model = LeadBoard
fields = ('id', 'event', 'score')
我添加了两名玩家
[
{
"id": 1,
"first_name": "Jhon",
"last_name": "Doe"
},
{
"id": 3,
"first_name": "Anna",
"last_name": "Doe"
}
]
及其在排行榜模型中的分数
[
{
"id": 1,
"player": 1,
"event": 1,
"score": 20
},
{
"id": 2,
"player": 3,
"event": 1,
"score": 90
}
]
这是排行榜的回应,
[
{
"id": 1,
"event": {
"id": 1,
"name": "event2020",
"user": [
{
"id": 1,
"first_name": "Jhon",
"last_name": "Doe"
},
{
"id": 3,
"first_name": "Anna",
"last_name": "Doe"
}
],
"desc": "event description"
},
"score": 20
},
{
"id": 2,
"event": {
"id": 1,
"name": "event2020",
"user": [
{
"id": 1,
"first_name": "Jhon",
"last_name": "Doe"
},
{
"id": 3,
"first_name": "Anna",
"last_name": "Doe"
}
],
"desc": "event description"
},
"score": 90
}
]
但我期望得到的是这样的响应,它正确地返回事件的玩家(用户)和他们的分数。
[
{
"id": 1,
"event": {
"id": 1,
"name": "event2020",
"user": [
{
"id": 1,
"first_name": "Jhon",
"last_name": "Doe",
"score": 20
},
{
"id": 3,
"first_name": "Anna",
"last_name": "Doe",
"score": 90
}
],
"desc": "event description"
}
}
]
我错过了什么?
我是Django和Django休息框架的新手。
1条答案
按热度按时间6jygbczu1#
你需要通过事件而不是排行榜做出React。
现在使用视图获取事件列表
更新1如果您想在玩家中获得分数
试试看。