在Rails中重构复杂的过滤器

Tal*_*boy 2 refactoring ruby-on-rails ruby-on-rails-4 rails-activerecord

我正在尝试处理一个有点复杂的查询.我已经阅读了一些关于如何处理这个问题的方法,但它们并没有真正适用于此,因为这不像一个复杂的搜索表单(如在vBulletin搜索帖表单上),而是一组过滤两者的路由按"类别"(未发布,流行,最新)和"时间"(所有时间,上个月,上周,今天)

我意识到下面的代码非常糟糕.我的目标只是让它工作,然后重构.更不用说,它甚至没有真正起作用,因为它没有考虑两种类型和时间,只考虑其中一种,但我想我会在这个线程中处理它.

另外,为了使这个SO代码粘贴更清楚,我.page(params[:page]).per(30)从每一行中排除了它,但是它需要继续所有这些.

那么,有谁知道我会怎么做呢?我已经仔细考虑了一段时间而且有点难过

def index
  case params[:category]
  when "latest"
    @books = Book.all.page(params[:page]).per(30)   
  when "downloads"
    @books = Book.order('downloads DESC')
  when "top100"
    @books = Book.order('downloads DESC').limit(100)
  when "unreleased"
    @books = Book.unreleased
  else
    @books = Book.all.page(params[:page]).per(30)   
  end

  case params[:time]
  when "today"
    @books = Book.days_old(1)
  when "week"
    @books = Book.days_old(7)
  when "month"
    @books = Book.days_old(30)
  when "all-time"
    @books = Book.all
  else
    @books = Book.all.page(params[:page]).per(30)   
  end
end  
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路线:

# Books
get 'book/:id', to: 'books#show', as: 'book'

resources :books, only: [:index] do
  get ':category/:time(/:page)', action: 'index', on: :collection
end
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dep*_*epa 7

  1. 将所有查询作为范围移动到模型

    class Book < ActiveRecord::Base
      scope :downloads,  -> { order('downloads DESC') }
      scope :top100,     -> { order('downloads DESC').limit(100) }
      scope :unreleased, -> { unreleased }
      scope :today,      -> { days_old(1) }
      scope :week,       -> { days_old(7) }
      scope :month,      -> { days_old(30) }
      scope :latest,     -> { }
      scope :all_time,   -> { }
    end
    
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  2. 创建辅助方法以过滤参数并避免不匹配的数据

    class BooksController < ApplicationController
      private
    
      def category_params
        %w(downloads top100 unreleased).include?(params[:category]) ? params[:category].to_sym : nil
      end
    
      def time_params
        %w(today week month latest all_time).include?(params[:time]) ? params[:time].to_sym : nil
      end
    end
    
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  3. case通过应用与params同名的范围来摆脱该语句

    def index
      query = Book.all
      query = query.send(category_params) if category_params
      query = query.send(time_params) if time_params
      @books = query.page(params[:page]).per(30)
    end
    
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在四条线上,我们仍然在Sandi Metz的指导范围内!:)