Question

我试图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，但它们并不适用于此，因为它不像一个复杂的搜索表单（比如在vBulletin搜索帖子表单中），而是一个按类别＆＃39;过滤的一组路线（未发行，流行，最新）和时间＆＃39; （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，因为它没有考虑到两种类型和时间，只是其中一种，但我想我会在这个线程中处理它。

另外，为了使这个SO代码粘贴更加清晰，我从每一行中排除了.page(params[:page]).per(30)，但是它需要继续使用它们。

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

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

Answer 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

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

通过应用与params同名的范围来删除case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

在四行，我们仍然在Sandi Metz' guidelines的范围内！：）

Answer 2

在rails中，您可以“链接”查询，例如

   Book.where(:released => true).where(:popular => true)

与

相同

   Book.where(:released => true, popular => true)

您可以使用它来帮助您进行重构。以下是我的看法：

   def index

      # Start with all books, we are going to add other filters later
      query = Book.scoped

      # Lets handle the time filter first   
      query = query.where(['created_at > ?', start_date] if start_date

      case params[:category]
      when "latest"
        query = query.order('created_at DESC')
      when "downloads"
        query = query.order('downloads DESC')
      when "top100"
        query = query.order('downloads DESC').limit(100)
      when "unreleased"
        query = query.where(:released => false)
      end

      # Finally, apply the paging
      @books = query.page(params[:page]).per(30)

end

    private

    def start_date
      case params[:time]
      when "today"
        1.day.ago 
      when "week"
        7.days.ago
      when "month"
        1.month.ago
      when "all-time"
        nil
      else
        nil
      end
    end

在Rails中重构复杂的过滤器

2 个答案: