ActiveRecord：具有“求和”，“合并”，“分组依据”和“选择多个字段”的复杂查询

我已经花了几天时间进行此查询，但仍然没有找到解决方案。

有我的模特

class UserObject < ActiveRecord::Base
  self.table_name = "user_objects"
  belongs_to :user, class_name: 'User'
  belongs_to :the_object, class_name: 'Object'
end

class Object < ActiveRecord::Base
  self.table_name = 'objects'

  has_many :user_objects, class_name: 'UserObject', foreign_key: :the_object_id, inverse_of: :the_object
  has_many :users, through: :user_objects
end

class User < ActiveRecord::Base
  self.table_name = "users"
end

这是我的图式

  create_table "objects", force: true do |t|
    t.float    "importance",        default: 0.5, null: false
  end

  create_table "user_objects", force: true do |t|
    t.integer  "user_id"
    t.integer  "the_object_id"
    t.float    "score",              default: 0.0,   null: false
  end

  create_table "users", force: true do |t|
    t.string   "first_name"
    t.string   "last_name"
  end

我需要选择objects.importance和总和的查询user_objects.score。但是我还必须选择仅查询objects属于user1和的那些查询user2。

我写了一个查询来查询选择对象。

Object.select("objects.importance").joins(:user_objects).
                                    where(user_objects: {user_id: [user1_id,user2_id]}).
                                    group(objects: :id).
                                    having('COUNT("user_objects"."id")=2')

并转换为

SELECT objects.importance FROM "objects" INNER JOIN "user_objects" ON "user_objects"."the_object_id" = "objects"."id" WHERE "user_objects"."user_id" IN (2, 7) GROUP BY "objects"."id" HAVING COUNT("user_objects"."id")=2

当我执行此查询时，我得到了这个响应

[#<Object id: nil, importance: 0.5>, #<Object id: nil, importance: 0.5>]

可以响应的对象数量可以。但是我仍然不知道如何计算这个查询总和user_objects.score。下一个查询不起作用

Object.select("objects.importance, SUM(user_objects.score)").
       joins(:user_objects).
       where(user_objects: {user_id: [user1_id,user2_id]}).
       group(objects: :id).having('COUNT("user_objects"."id")=2')

我期望这样的回应

[#[<Object id: nil, importance: 0.5>, 0.2], #[<Object id: nil, importance: 0.5>,0.3]]

我非常感谢您在解决此问题方面可以给我的任何帮助。