我有以下模型:
class User(models.Model):
name = models.Charfield()
email = models.EmailField()
class Friendship(models.Model):
from_friend = models.ForeignKey(User)
to_friend = models.ForeignKey(User)
这些模型在以下视图和序列化器中使用:
class GetAllUsers(generics.ListAPIView):
authentication_classes = (SessionAuthentication, TokenAuthentication)
permission_classes = (permissions.IsAuthenticated,)
serializer_class = GetAllUsersSerializer
model = User
def get_queryset(self):
return User.objects.all()
class GetAllUsersSerializer(serializers.ModelSerializer):
is_friend_already = serializers.SerializerMethodField('get_is_friend_already')
class Meta:
model = User
fields = ('id', 'name', 'email', 'is_friend_already',)
def get_is_friend_already(self, obj):
request = self.context.get('request', None)
if request.user != obj and Friendship.objects.filter(from_friend = user):
return True
else:
return False
基本上,对于由GetAllUsers
视图返回的每个用户,我想打印出该用户是否与请求者是朋友(实际上我应该检查from_ 和to_friend,但这并不影响问题的关键)。
我发现,在数据库中存在N个用户时,有1个查询用于获取所有N个用户,然后在序列化器的get_is_friend_already
中进行1xN次查询。
是否有一种避免这种情况的rest-framework方式?也许像传递一个包含相关Friendship
行的select_related
查询给序列化器这样的东西?