Rails通过连接表获得关联计数

max*_*max 7 postgresql performance ruby-on-rails

这个问题是Rails中HABTM协会的一个分支:收集和计算模型儿童的类别.

鉴于:

class Category < ActiveRecord::Base
  has_and_belongs_to_many :books
  validates_uniqueness_of :name
end

class Book < ActiveRecord::Base
  has_and_belongs_to_many :categories
end

class Store < ActiveRecord::Base
  has_many :books
  has_many :categories, through: :books
end
Run Code Online (Sandbox Code Playgroud)

任务:

给定商店,列出每个类别的书籍数量.

Store.first.books_per_category
Run Code Online (Sandbox Code Playgroud)

期望的输出:

[ { name: 'mystery', count: 5 }, { name: 'fantasy', count: 6 } ]
Run Code Online (Sandbox Code Playgroud)

然而,每家商店可能拥有大量的书籍和类别.

我正在尝试创建一个单一的高性能查询,该查询只获取名称列和与商店关联的每个不同类别的书籍计数,而不将书籍加载到内存中.

我到目前为止尝试过:

class Store < ActiveRecord::Base

  # Will load each book into memory
  def books_per_category
    categories.eager_load(:books).map do |c|
      {
          name: c.name,
          count: c.books.size # Using size instead of count is important since count will always query the DB
      }
    end
  end

  # will query books count for each category.
  def books_per_category2
    categories.distinct.map do |c|
      {
          name: c.name,
          count: c.books.count
      }
    end
  end
end
Run Code Online (Sandbox Code Playgroud)

数据库架构:

ActiveRecord::Schema.define(version: 20150508184514) do

  create_table "books", force: true do |t|
    t.string   "title"
    t.datetime "created_at"
    t.datetime "updated_at"
    t.integer  "store_id"
  end

  add_index "books", ["store_id"], name: "index_books_on_store_id"

  create_table "books_categories", id: false, force: true do |t|
    t.integer "book_id",     null: false
    t.integer "category_id", null: false
  end

  add_index "books_categories", ["book_id", "category_id"], name: "index_books_categories_on_book_id_and_category_id"
  add_index "books_categories", ["category_id", "book_id"], name: "index_books_categories_on_category_id_and_book_id"

  create_table "categories", force: true do |t|
    t.string   "name"
    t.datetime "created_at"
    t.datetime "updated_at"
  end

  create_table "stores", force: true do |t|
    t.string   "name"
    t.datetime "created_at"
    t.datetime "updated_at"
  end
end
Run Code Online (Sandbox Code Playgroud)

Jak*_*ski 7

您可以使用链selectgroup汇总每个类别的图书数量.您的books_per_category方法可能如下所示:

def books_per_category
  categories.select('categories.id, categories.name, count(books.id) as count').group('categories.id, categories.name').map do |c|
    {
      name: c.name,
      count: c.count
    }
  end
end
Run Code Online (Sandbox Code Playgroud)

这将产生以下SQL查询:

SELECT categories.id, categories.name, count(books.id) as count 
  FROM "categories" 
  INNER JOIN "books_categories" ON "categories"."id" = "books_categories"."category_id" 
  INNER JOIN "books" ON "books_categories"."book_id" = "books"."id" 
  WHERE "books"."store_id" = 1 
  GROUP BY categories.id, categories.name
Run Code Online (Sandbox Code Playgroud)


Phi*_*arg 6

您将希望在 Categories 对象上创建一个方法(或范围),例如。

Category.joins(:books).select('categories.*, COUNT(books.id) as book_count').group('categories.id')
Run Code Online (Sandbox Code Playgroud)

结果对象现在将具有类别实例的每个属性,并响应一个方法,book_count该方法返回具有该实例类别 ID 的书籍数量。值得一提的是,这将省略任何没有与其关联的书籍的类别。如果要包含这些,则需要将查询更新为以下内容:

Category.joins('LEFT OUTER JOIN books_categories on books_categories.category_id = categories.id').select('categories.*, COUNT(books_categories.book_id) as book_count').group('categories.id')
Run Code Online (Sandbox Code Playgroud)