Django-过滤相关对象

Question

我有一个难题。

这些是我的型号：

class StatusGroup(models.Model):
name = models.TextField()
def __str__(self):
return self.name

class StatusDetail(models.Model):
action = models.CharField(choices=[("CORRECT", "CORRECT"),
("INCORRECT", "INCORRECT")],
max_length=64)
status_group = models.ForeignKey(to=StatusGroup,
on_delete=models.CASCADE,
related_name="status_details")
def __str__(self):
return f"Detail: {self.action}"

序列化程序：

class StatusDetailSerializer(serializers.ModelSerializer):
class Meta:
model= models.StatusDetail
fields = "__all__"
class StatusGroupSerializer(serializers.ModelSerializer):
status_details = StatusDetailSerializer(many=True)
class Meta:
model = models.StatusGroup
fields = [
"pk",
"status_details",
"name"
]

和一个观点：

class Status(viewsets.ModelViewSet):
queryset = models.StatusGroup.objects.all()
serializer_class = serializers.StatusGroupSerializer
authentication_classes = []
permission_classes = [permissions.AllowAny]
filter_backends = (DjangoFilterBackend,)
filterset_fields = ['status_details__action']

当我击中localhost:8000/api/status?status_details__action=INCORRECT时

我得到：

[
{
"pk": 2,
"status_details": [
{
"id": 3,
"action": "CORRECT",
"status_group": 2
},
{
"id": 4,
"action": "INCORRECT",
"status_group": 2
}
],
"name": "Mixed"
}
]

而我想要：

[
{
"pk": 2,
"status_details": [
{
"id": 4,
"action": "INCORRECT",
"status_group": 2
}
],
"name": "Mixed"
}
]

如何强制Django过滤相关对象？我可以在SQL控制台中获得我想要的结果，但Django添加了所有属于StatusGroup的相关对象。我有一个误解，但我不知道那是什么。

Answer 1

尝试这种方式(使用查询集(：

from django.db.models import Prefetch
queryset = StatusGroup.objects.prefetch_related(
Prefetch('status_details', queryset=StatusDetail.objects.filter(action='what_you_want'), to_attr='action'))

并且在您的序列化程序类中：

class StatusGroupSerializer(serializers.ModelSerializer):
status_details = StatusDetailSerializer(source='action', many=True, read_only=True)

Answer 2

首先可以创建一个FilterSet类，如下所示：

from django_filters import rest_framework as filters
class StatusGroupFilter(filters.FilterSet):
action = filters.CharFilter(
field_name="status_details__action",
lookup_expr="iexact"
)
class Meta:
model = StatusGroup
fields = ["action"]

那么在你看来：

class Status(viewsets.ModelViewSet):
queryset = models.StatusGroup.objects.all()
serializer_class = serializers.StatusGroupSerializer
authentication_classes = []
permission_classes = [permissions.AllowAny]
filter_backends = (DjangoFilterBackend,)
# Set the filterset class here
filterset_class = StatusGroupFilter

然后在URL中，您可以调用localhost:8000/api/status?action=INCORRECT