如何重构 ActiveRecord 查询？

Question

我在控制器中有一个代码：

def latest
  @latest_articles = user_signed_in? ? Article.limit(10).order(id: :desc).pluck(:id, :title) : Article.where("status = ?", Article.statuses[:public_article]).limit(10).order(id: :desc).pluck(:id, :title)
  render json: @latest_articles
end

如何重构让它看起来更优雅？我尝试使用 lambda：

extract = lambda {|a| a.order(id: :desc).pluck(:id, :title)}
Article.limit(10) {|a| a.extract}

但它 returns 仅 Article.limit(10)

UPD：如果用户已登录，我需要获取所有文章的最后 10 篇，如果未登录，则仅获取 public 文章的最后 10 篇。

Answer 1

我会创建一个初始范围，并根据某些条件对其进行修改：

def latest
  scope = Article.order(id: :desc)
  scope = scope.where(status: Article.statuses[:public_article]) if user_signed_in?

  render json: scope.limit(10).pluck(:id, :title)
end

Answer 2

您可以重构为

@lates_articles = Article.all
@lates_articles = @latest_articles.where("status = ?", Article.statuses[:public_article]) unless user_signed_in?
render json: @latest_articles.limit(10).order(id: :desc).pluck(:id, :title)

但是创建模型方法会更好

class Article < ActiveRecord::Base
  ...
  scope :latest, -> {last(10).order(id: :desc)}

  def self.public user_signed
    if user_signed
      all
    else
      where("status = ?", statuses[:public_article])
    end
  end
  ...
end

然后你会像

那样使用它

def latest
  render json: Article.public(user_signed_in?).latest.pluck(:id, :title)
end

Answer 3

最终版本：

def latest
  scope = Article.order(id: :desc)
  scope = scope.shared unless user_signed_in?
  render json: scope.limit(10), except: [:body, :created_at, :updated_at]
end

如何重构 ActiveRecord 查询？

How to refactor ActiveRecord query?

refactoring

activerecord

ruby-on-rails