Rails：通过连接表获取关联计数

Question

我有一个名为“Fact”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“Feedback >" 有 3 个字段 [user_id, fact_id, score] 每次一个人给出反馈时，他得到的反馈是谁做的（user_id）到什么事实（fact_id）和什么是合格的（0,1,2）。 好吧，现在我想获取包含其资格的事实列表，例如：

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

其中 number_of 是符合条件的次数。

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 以便同时返回反馈为 0 的事实（每个计数列均返回 0）。

如Rohan所说，您需要在fact.rb中添加has_many :feedbacks

有更复杂的解决方案可以适应任意数量的可能分数，但在这种情况下，这将是过分热心的。

编辑：对于 SQLite（我认为它支持相关子查询...）

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

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

Answer 3

根据帖子中提供的描述，您似乎需要事实详细信息以及每个 uniq 事实的分数。

下面提到的一些东西可以帮助你实现：

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")

上面的查询是将反馈与事实结合起来，然后对得分值进行分组，因为只有三个值，即 0、1、2（如帖子中所述）。