Ruby on rails “不明白为什么”；预加载/预加载；在ROR中不工作？

我不明白为什么“域”被多次查询，尽管我在控制器中使用了渴望的加载（像服务器，但“服务器”只被查询了一次）

代码：

在控制器中：

       @problems = Problem.all.eager_load(:serverity, :domain, :nation,  :status,:urgency,:user, :works).order(start_time: :desc)

在模型中：

    class Problem < ActiveRecord::Base
     belongs_to :serverity
     belongs_to :urgency
     belongs_to :nation
     belongs_to :status
     belongs_to :user  
     belongs_to :cause
     belongs_to :domain
     has_many   :works
     has_many   :attachments, :dependent => :destroy   
   end

---

请显示

问题的代码

。什么是

domian.parent

？我已更新了问题。领域就是祖先。域1->域2->域3->域4。domain4.parent是domain3，依此类推。这些是您的查询。你可以做

{:domain=>:parent}

，但是你可能有多个祖先这只会对父母有帮助。我补充说，但是父母/孩子是祖先的方法，而不是关联：（.如何预加载孩子。

<% @problems.each do |problem| %>
       <td>
    <%= link_to problem.name, problem, :method => :get %></td>
    <td>
      <% if problem.domain.root? %>
      <%= problem.domain.name %>
      <%else%>
      <%= problem.domain.parent.name %>
      <%end%>
    </td> 
    <td><%= problem.serverity.name %></td>
<%>

  Processing by ProblemsController#index as HTML                                                                           
  [1m[36mUser Load (15.6ms)[0m  [1m[34mSELECT  `users`.* FROM `users` WHERE          `users`.`id` = 1 LIMIT 1[0m              
  [1m[36mCACHE (0.0ms)[0m  [1m[34mSELECT  `users`.* FROM `users` WHERE `users`.`id` = 1 LIMIT 1[0m  [["id", 1], ["LIM
  Rendering problems/index.html.erb within layouts/application                                                             
  [1m[35m (78.0ms)[0m  [1m[34mSELECT COUNT(DISTINCT `problems`.`id`) FROM `problems` LEFT OUTER JOIN `serverities` ON 
  [1m[35m (0.0ms)[0m  [1m[34mSELECT COUNT(DISTINCT `problems`.`id`) FROM `problems` LEFT OUTER JOIN `serverities` ON `
  [1m[35m (0.0ms)[0m  [1m[34mSELECT COUNT(DISTINCT `problems`.`id`) FROM `problems` LEFT OUTER JOIN `serverities` ON `
  [1m[35m (0.0ms)[0m  [1m[34mSELECT COUNT(DISTINCT `problems`.`id`) FROM `problems` LEFT OUTER JOIN `serverities` ON `
  [1m[35m (0.0ms)[0m  [1m[34mSELECT COUNT(DISTINCT `problems`.`id`) FROM `problems` LEFT OUTER JOIN `serverities` ON `
  [1m[35m (0.0ms)[0m  [1m[34mSELECT COUNT(DISTINCT `problems`.`id`) FROM `problems` LEFT OUTER JOIN `serverities` ON `
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