Question

我有一个名为“ 事实”的实体，该实体与另一个名为“ User ”的实体有关系，这种关系称为“ 反馈”，该字段具有3个字段， [user_id，fact_id，分数] ，每当有人提供反馈时，他都会得到对事实做出的反馈（ user_id ）（< strong> fact_id ）和符合条件的内容（ 0,1,2 ）。好吧，现在我想获取包含其资格的事实清单，例如：

fact : {id: 3, name: "some name", number_of_zero: 7, number_of_one: 3, number_of_two: 3}

其中number_of是合格的次数。

数据库架构：

  create_table "facts", force: :cascade do |t|
    t.string "title"
    t.string "description"
    t.integer "user_id"
  end

  create_table "users", force: :cascade do |t|
    t.string "name"
    t.string "last_name"
  end

create_table "feedbacks", force: :cascade do |t|
    t.integer "score"
    t.integer "user_id"
    t.integer "fact_id"
    t.index ["fact_id"], name: "index_feedbacks_on_fact_id"
    t.index ["user_id"], name: "index_feedbacks_on_user_id"
  end

关系：

class Feedback < ApplicationRecord
  belongs_to :user
  belongs_to :fact
end

Answer 1

Fact.
  group(:id, :name).
  select(:id, name).
  left_joins(:feedbacks).
  select("COUNT(feedbacks.id) FILTER (WHERE score = 0) AS number_of_zero").
  select("COUNT(feedbacks.id) FILTER (WHERE score = 1) AS number_of_one").
  select("COUNT(feedbacks.id) FILTER (WHERE score = 2) AS number_of_two")

您需要left_joins，以便还返回反馈为0的事实（每个计数列为0）。

如罗汉所言，您需要在has_many :feedbacks中添加fact.rb

有更复杂的解决方案可以适应任何数量的可能得分，但在这种情况下，可能会显得过于热情。

编辑：对于SQLite （我认为它支持相关的子查询...）

Fact.
  select(:id, name).
  select("(SELECT COUNT(*) FROM feedbacks WHERE score = 0 AND fact_id = facts.id) AS number_of_zero").
  select("(SELECT COUNT(*) FROM feedbacks WHERE score = 1 AND fact_id = facts.id) AS number_of_one").
  select("(SELECT COUNT(*) FROM feedbacks WHERE score = 2 AND fact_id = facts.id) AS number_of_two")

Answer 2

根据帖子中提供的描述，您似乎想要事实的详细信息以及每个uniq事实的得分。

下面提到的某些方法可以帮助您实现：

Fact.joins(:feedbacks).select("facts.id, facts.title,CASE when feedbacks.score is 0 
then count(feedbacks.score) as score_0 end,
CASE when feedbacks.score is 1 then count(feedbacks.score) as score_1 end,
CASE when feedbacks.score is 2 then count(feedbacks.score) as score_2 end")
.group("feedbacks.score")

上述查询将反馈与事实结合在一起，然后对得分值进行分组，因为只有三个值，即0,1,2（如文章中所述）。

Answer 3

Model:
class Feedback < ApplicationRecord
  belongs_to :user
  belongs_to :fact
end
class Fact < ApplicationRecord
 has_many :feedbacks
end
 fact = Fact.all
  array = []
 fact.each do |each_fact|
    arr = {}
    each_fact.feedbacks do |each_feedback|
       arr["id"] = each_feed_back.id
       arr["name"] = each_feedback.name
       arr["no_of_zero"] = each_feedback.score 
       array << arr
   end
end

Rails：通过连接表获取关联计数

3 个答案: