相关疑难解决方法(0)

我有以下型号:

class User(models.Model):
    name = models.Charfield()
    email = models.EmailField()

class Friendship(models.Model):
    from_friend = models.ForeignKey(User)
    to_friend = models.ForeignKey(User)

这些模型用于以下视图和序列化程序:

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

所以基本上,对于GetAllUsers视图返回的每个用户,我想打印出用户是否是请求者的朋友(实际上我应该检查from_和to_friend,但对于问题并不重要)

我看到的是,对于数据库中的N个用户,有1个查询用于获取所有N个用户,然后在序列化程序中查询1xN个查询 get_is_friend_already

有没有办法在休息框架方式中避免这种情况？也许是将select_related …

序幕:

这是SO中经常出现的问题:

• Django模型组
• Django相当于count和group by
• 如何在django中查询GROUP BY？
• 如何使用ORM相当于SQL计数,组和连接查询？

我已经在SO文档中编写了一个示例,但由于文档将于2017年8月8日关闭,我将遵循这个广泛上升和讨论的元答案的建议,并将我的示例转换为自我回答的帖子.

当然,我也很乐意看到任何不同的方法!

题:

假设模型:

class Books(models.Model):
    title  = models.CharField()
    author = models.CharField()
    price = models.FloatField()

如何使用Django ORM在该模型上执行以下查询:

• GROUP BY ... COUNT:

  SELECT author, COUNT(author) AS count
  FROM myapp_books GROUP BY author
  

  Run Code Online (Sandbox Code Playgroud)

• GROUP BY ... SUM:

  SELECT author,  SUM (price) AS total_price
  FROM myapp_books GROUP BY author
  

  Run Code Online (Sandbox Code Playgroud)