【问题标题】:Devise polymorphic association nested attributes with simple form using rails 4使用rails 4设计具有简单形式的多态关联嵌套属性
【发布时间】:2013-09-06 20:21:19
【问题描述】:

我正在与设计和简单的多态关联,但由于某种原因我无法让参数工作

这是我的代码:

用户:

class User < ActiveRecord::Base
  devise :database_authenticatable,
         :rememberable, :trackable, :validatable

  belongs_to :loginable, polymorphic: true
end

设计师:

class Designer < ActiveRecord::Base
  has_one :user, as: :loginable
  accepts_nested_attributes_for :user

end

布局:

<%= simple_form_for [:admin, @designer] , :html => { :class => 'form-horizontal' } do |f| %>
    <% if f.error_notification %>
        <div class="alert alert-error fade in">
          <a class="close" data-dismiss="alert" href="#">&times;</a>
          <%= f.error_notification %>
          <% if @designer.errors.any? %>
              <ul>
                <% @designer.errors.full_messages.each do |msg| %>
                    <li><%= msg %></li>
                <% end %>
              </ul>
          <% end %>

        </div>
    <% end %>

    <div class="control-group">
      <%= f.label :profile_name, :class => 'control-label' %>
      <div class="controls">
        <%= f.text_field :profile_name, :class => 'text_field' %>
      </div>
    </div>

    <%= f.simple_fields_for users do |u| %>

        <div class="control-group">
          <%= u.label :email, :class => 'control-label' %>
          <div class="controls">
            <%= u.text_field :email, :class => 'text_field' %>
          </div>
        </div>

        <div class="control-group">
          <%= u.label :password, :class => 'control-label' %>
          <div class="controls">
            <%= u.password_field :password, :class => 'text_field' %>
          </div>
        </div>

        <div class="control-group">
          <%= u.label :type, :class => 'control-label' %>
          <div class="controls">
            <%= u.input :role, :label => false do %>
                <%= u.select :role_id, Role.all.map { |r| [r.name, r.id] } %>
            <% end %>
          </div>
        </div>

        <div class="control-group">
          <%= u.label :firstname, :class => 'control-label' %>
          <div class="controls">
            <%= u.text_field :firstname, :class => 'text_field' %>
          </div>
        </div>
    <% end %>


    <div class="form-actions">
      <%= f.submit nil, :class => 'btn btn-primary' %>
    </div>
<% end %>

控制器:

  def create
    @user = User.new(designer_params)
    @designer = Designer.new(designer_params)
    @user.loginable = @designer

    respond_to do |format|
      if @user.save! && @designer.save!
        format.html { redirect_to admin_designers_path, notice: 'Designer was successfully created.' }
        format.json { render action: 'show', status: :created, location: admin_designer_path(@designer) }
      else
        format.html { render action: 'new' }
        format.json { render json: [designer: @designer.errors, user: @user.errors], status: :unprocessable_entity }
      end
    end
  end

  def designer_params
    params.permit(:profile_name, :user, user_attributes: [:email, :password, :password, :firstname, :lastname, :address, :postalcode, :city, :country, :role, :role_id])
  end

我的参数似乎忽略了用户属性,出于某种原因我只看到了个人资料名称。

任何关于如何解决此问题的想法将不胜感激

谢谢!

【问题讨论】:

    标签: devise ruby-on-rails-4 nested-attributes polymorphic-associations


    【解决方案1】:

    我设法解决了自己的问题。所以我发布答案希望可以节省其他人很多时间。

    我最终为用户参数创建了另一个私有方法

      def designer_params
        params.require(:designer).permit(:profile_name, user_attributes: [:email, :password, :password, :firstname, :lastname, :address, :postalcode, :city, :country, :role, :role_id])
      end
      def user_params
        params[:designer][:user_attributes].permit(:email, :password, :password, :firstname, :lastname, :address, :postalcode, :city, :country, :role, :role_id)
      end
    

    然后用这些来建立我的关系

      def create
        @designer = Designer.new(designer_params)
        @user = User.new(user_params)
        @user.loginable = @designer
        @designer.save!
      end
    

    如果您在查看嵌套表单时遇到问题,请务必使用 build_方法

      def new
        @designer = Designer.new
        @user = @designer.build_user
      end
    

    【讨论】:

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