如何通过模型动作对红宝石数组进行排序？

Question

I'm trying to make a project like cafe search service based on social network, and I want to sort Cafe array by points which other users gave.

class Cafe < ActiveRecord::Base

  belongs_to :user

  has_many :posts, dependent: :destroy
  has_many :tags, dependent: :destroy
  has_many :payments, dependent: :destroy
  has_many :payinfos, dependent: :destroy
  mount_uploader :image,CafeimageuploaderUploader
  mount_uploader :thumnail,CafeimageuploaderUploader

  geocoded_by :address
  after_validation :geocode
  def avg
    total = 0
    posts.each do |c|
      total += c.score
    end
    if posts.count == 0
        0
    else
      total.to_f / posts.count
    end
  end
end

this is Cafe model, 'avg' is point average that users gave.

class CreatePosts < ActiveRecord::Migration
  def change
    create_table :posts do |t|

      t.string :content              , null: false, default: ""
      t.string :image
      t.string :address              , null: false, default: "위치정보 없음"
      t.string :hashstr
      t.datetime :writtentime
      t.integer :user_id
      t.integer :cafe_id             , null: false, default: 0
      t.integer :score, default:0
      t.timestamps null: false
    end
  end
end

Here's post columns.

What I want to do is sort new Cafe array by this avg action.(is it called model action, right??)

give me some advice, thank you.

Answer 1

I would add an attribute average_score to cafe model and a after_save callback method to post model. If a post will be safed you take all posts of a cafe (by the foreign key), calculate the average score and save it to the cafe or you trigger a method of the cafe model to do that. So you can sort your cafes easily.

Answer 2

Let use db query, it is more efficient. (assume Cafe's table name is cafes )

Cafe.joins("LEFT OUTER JOIN (SELECT cafe_id, AVG(score) AS avg_score
                             FROM posts
                             GROUP BY cafe_id
                            ) AS temp
            ON cafes.id = temp.cafe_id
          ").order('temp.avg_score DESC NULLS LAST')

Ideally, we calculate the average score in a temporary table first then join with Cafe via cafe_id . After that, we can easily use order with avg_score . Because we are using LEFT OUTER JOIN , the avg_score of cafe which doesn't have any post will be NULL , so we use NULLS LAST to ensure that cafe will be in the end.