ActiveAdmin -- 如何显示类别分类？（在树类型层次结构中）

发布于 2024-12-15 00:30:53 字数 358 浏览 3 评论 0原文

我有一个模型 category ，其中包含一个名为 category.parent_id 的字段，用于创建分类法（带有子类别的顶级类别）。

目前，我的索引页面显示的类别为：

Top level category
Sub category
Sub category
Top category

我如何以正确的分类法对它们进行排序（即首先所有顶级类别均取自数据库，然后是子类别等）并显示如下：

Top level category
-- Sub category
-- Sub category
Top category

原文

I have a model category which contains a field called category.parent_id which is used to create a taxonomy (top level categories with sub categories).

At the moment my index page shows the categories as:

Top level category
Sub category
Sub category
Top category

How could I order them in a proper taxonomy (i.e. first all top level categories are taken from db then sub categories, etc) and shown as the following:

Top level category
-- Sub category
-- Sub category
Top category

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旧人九事 2024-12-22 00:30:53

经过几个小时的头撞墙后，我想出了这个解决方案：

ActiveAdmin 无法无法渲染整个索引的一部分，但是您可以为单个列渲染一个！

因此，解决方法是为子类别创建一个列，如下所示：

    column "Sub Categories" do |category|
      children = Category.children(category.id)
      if children.present?
        render :partial => "children", :locals => { :children => children }
      end
    end

然后在 _children.html.erb 视图部分中，您可以打印出如下列表：

    <% children.each do |child| %>
        <p><%= link_to child.name, edit_admin_category_path(child) %></p>
    <% end %>

After hours of banging my head against the wall, I came up with this solution:

ActiveAdmin can not render a partial for the entire index, however you can render one for a single column!

So the work around was to create a column for sub categories like this:

    column "Sub Categories" do |category|
      children = Category.children(category.id)
      if children.present?
        render :partial => "children", :locals => { :children => children }
      end
    end

And then in _children.html.erb view partial you can print out the list like this:

    <% children.each do |child| %>
        <p><%= link_to child.name, edit_admin_category_path(child) %></p>
    <% end %>

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鸢与 2024-12-22 00:30:53

在您的情况下，您可以创建一个像这样的自引用模型（在 app/model/category.rb 中）：

belongs_to :parent, :class_name => 'Category', :foreign_key => 'parent_id'
has_many :children, :class_name => 'Category', :foreign_key => 'parent_id'

然后您可以为父项创建一个范围

scope :parents, where("parent_id IS NULL")

，并且在视图中您可以使用这样的迭代

<ul>
<% Category.all.parents.each do |parent| %>
     <li class="parent"><%= link_to parent.name, parent %></li>
     <% parent.children.each do |child| %>
          <li class="sub"><%= link_to child.name, child %></li>
     <% end %>
<% end %>
</ul>

希望这样对你有帮助！

//我读了我想读的内容，...我没有看到它是关于主动管理的。以为是关于 ActiveRecord -.-

You could create a self referencing Model like this in you case (In app/model/category.rb):

belongs_to :parent, :class_name => 'Category', :foreign_key => 'parent_id'
has_many :children, :class_name => 'Category', :foreign_key => 'parent_id'

Then you can create a scope for the parents

scope :parents, where("parent_id IS NULL")

And in view you can use iteration like this

<ul>
<% Category.all.parents.each do |parent| %>
     <li class="parent"><%= link_to parent.name, parent %></li>
     <% parent.children.each do |child| %>
          <li class="sub"><%= link_to child.name, child %></li>
     <% end %>
<% end %>
</ul>

Hope this helps you!

//I read what I wanted to read,... I didn't see it's about active admin. Thought it's about ActiveRecord -.-

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