Rails 5:显示对象值而不是ID

Question

我有一款简单的跳伞应用程序。降落伞包装工跟踪他们的包装工作。每个降落伞钻机有几个包件。

我通过在平台细节(即反射#2)下显示rig_type_number，使显示视图工作得很好，但我只能让rig_id在索引的同一列中工作。理想情况下，我也希望使用rig_type_number而不是ID。

如何在视图中引用@rigs.rig_type_number?的效果？

这对我来说是有道理的，但不管用：

<td><%= link_to packjob.rig_id,rig_path(rig.rig_type_number) %>

我必须为索引控制器设置where子句吗？

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db：

class CreatePackjobs < ActiveRecord::Migration[5.1]
  def change
    create_table :packjobs do |t|
      t.string :packer
      t.string :rig
      t.references :rig, foreign_key: true

      t.timestamps
    end

  end
end

class CreateRigs < ActiveRecord::Migration[5.1]
  def change
    create_table :rigs do |t|
      t.boolean :rig_status
      t.string :rig_type_number
      t.integer :rig_season_jumpnum
      t.date :rig_res_last
      t.string :rig_res_who

      t.timestamps
    end
  end
end

packjobs_controller：

class PackjobsController < ApplicationController
  def index
    @packjobs = Packjob.paginate(page: params[:page]).order('id DESC') 
    @rigs = Rig.where(rig_status: "t")
  end

  def show
    @packjob = Packjob.find(params[:id])
    @rig = Rig.find(@packjob.rig_id)
  end

  def new
    @packjob = Packjob.new
    @rigs = Rig.where(rig_status: "t")
  end

  def edit
    @packjob = Packjob.find(params[:id])
    @rigs = Rig.where(rig_status: "t")
  end

  def create
    @packjob = Packjob.new(packjob_params)
    @rigs = Rig.where(rig_status: "t")
    if @packjob.save
      redirect_to @packjob
    else
      render 'new'
    end
  end


  def update
    @packjob = Packjob.find(params[:id])
    @rigs = Rig.where(rig_status: "t")
    if @packjob.update(packjob_params)
      redirect_to @packjob
    else
      render 'edit'
    end
  end


  def destroy
    @packjob = Packjob.find(params[:id])
    @packjob.destroy

    redirect_to packjobs_path
  end

  private
    def packjob_params
      params.require(:packjob).permit(:packer, :rig_id)
    end

    def rigs_params
      params.require(:rig).permit(:rig_status, :rig_type_number)
    end


end

show.html.erb：

<p>
   <b>Datestamp:</b>
   <%= @packjob.created_at.strftime("%m/%d/%Y %I:%M %P") %><br>
</p>
<p>
  <strong>Packer:</strong>
  <%= @packjob.packer %>
</p>

<p>
  <strong>Rig:</strong>
  <%= @rig.rig_type_number %><br>
</p>

<%= link_to 'Edit', edit_packjob_path(@packjob) %> |
<%= link_to 'Back', packjobs_path %>

index.html.erb：

<h1>Listing Packjobs</h1>

<%= link_to 'New Packjob', new_packjob_path %> | <br>

<br>

<div class="digg_pagination">
  <%= will_paginate @packjobs, :container => false %>
</div>

<table id=listtable CELLPADDING="4" border="1">
  <tr class="headerBlue" >
    <th>Datestamp</th>
    <th>Packer</th>
    <th>Rig</th>
    <th colspan="3"></th>
  </tr>

  <% @packjobs.each do |packjob| %>
  <tr class="<%= cycle('rowA', 'rowB') %>">
        <td><%= link_to packjob.created_at.strftime("%m/%d/%Y %I:%M:%S %P"),packjob_path(packjob.id) %></td>
        <td><%= packjob.packer %></td>
        <td><%= link_to packjob.rig_id,rig_path(packjob.rig_id) %>
        <td><%= link_to 'Show', packjob_path(packjob) %></td>
        <td><%= link_to 'Edit', edit_packjob_path(packjob) %></td>
        <td><%= link_to 'Destroy', packjob_path(packjob),
          method: :delete,
          data: { confirm: 'Are you sure?' } %></td>

    <% end %>
  </tr>
</table>
<div class="digg_pagination">
  <div class="page_info">
    <%= page_entries_info @packjobs %>
  </div>
  <%= will_paginate @packjobs,  :container => false %>
</div>

<br><%= link_to 'New Packjob', new_packjob_path %> |

编辑：

模型：

class Packjob < ApplicationRecord

  belongs_to :rig
  validates :packer, presence: true
  validates :rig_id, presence: true

  self.per_page = 25
end

class Rig < ApplicationRecord
  has_many :packjobs
  validates :rig_type_number, presence: true
  self.per_page = 25

end

Answer 1

如果正确设置了关联，则可以引用以下关联对象：

<td><%= link_to packjob.rig.rig_type_number,rig_path(packjob.rig_id) %>

Answer 2

您可以按下面的方式调用它，以防在PackJob模型中有PackJob关联：

<td><%= link_to packjob..rig.rig_type_number,rig_path(packjob.rig_id) if packjob.rig.present? %> </td>