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如何重构这个RubyonRails代码?

dry refactoring ruby-on-rails

Yuval Karmi · 技术社区 · 14 年前

我想根据职位的状态获取职位,所以我在我的 PostsController index 行动。不过,它似乎在混乱索引操作,我不确定它是否属于这里。

我怎样才能使它更简洁,在我的应用程序中将它移动到哪里,这样它就不会扰乱我的索引操作(如果这是正确的操作)?

if params[:status].empty?
  status = 'active'
else
  status = ['active', 'deleted', 'commented'].include?(params[:status]) ? params[:status] : 'active'
end

case status
when 'active'
  #active posts are not marked as deleted and have no comments
  is_deleted = false
  comments_count_sign = "="
when 'deleted'
  #deleted posts are marked as deleted and have no comments
  is_deleted = true
  comments_count_sign = "="
when 'commented'
  #commented posts are not marked as deleted and do have comments
  is_deleted = false
  comments_count_sign = ">"
end

@posts = Post.find(:all, :conditions => ["is_deleted = ? and comments_count_sign #{comments_count_sign} 0", is_deleted])

2 回复 | 直到 14 年前

zed_0xff 14 年前

class Post < ActiveRecord::Base
  named_scope :active, :conditions => { :is_deleted => false, :emails_count => 0 }
  named_scope :sent, :conditions => ["is_deleted = ? AND emails_count > 0", true]
  ...
end

使用它,如post.active.all、post.active.first、post.active.each等

然后

status = %w'active deleted sent'.include?(params[:status]) : params[:status] : 'active'
@posts = Post.send(status).all

GSP 14 年前

我会考虑在post中添加一个类方法

def file_all_based_on_status status

  # custom logic for queries based on the given status here
  # handling nils and other cases

end

这样你的控制器索引就简单了

def index
  @posts = Post.find_all_based_on_status params[:status]
end

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