使用friendly_id匹配单个Rails路由中的多个模型

Question

我有一个 Company 和一个 User 模型，两者都通过 friendly_id 使用 slug。确保 Slug 在两种模型中都是唯一的。

我想要网址：

http://www.example.com/any_company_name

http://www.example.com/any_user_name

例如/apple 和 /tim

我不确定如何在 Rails 中实现这一点。

我尝试了以下各种排列：

routes.rb:
  resources :users, path: ''
  resources :companies, path: ''
  get '*search', to: 'my_controller#redirect'

和

my_controller#redirect:
  @company = Company.friendly.find(params[:search])
  redirect_to @company if @company
  @user = User.friendly.find(params[:search])
  redirect_to @user if @user

但是我无法让它工作。我可以让/apple 重定向到/companies/apple 和/tim 重定向到/users/tim（通过删除path: '' 选项）但这不是我想要实现的。

Answer 1

您可以使用 url_for 使用 controller#action 进行重定向，而不是使用 url 进行重定向。

例如：

my_controller#redirect:
  @company = Company.friendly.find(params[:search])
  redirect_to url_for(action: 'show', controller: :companies , status: :success, company_id:@company.id)
  @user = User.friendly.find(params[:search])
  redirect_to url_for(action: 'show', controller: :users, status: :success,user_id:@user.id)

Answer 2

我遇到了类似的问题，并且能够通过创建具有 slug 属性和与“Sluggable”类的多态关联的 PublicSlug 模型来解决它。我还使用了我想要查询的模型中包含的 Sluggable 关注点。

PublicSlug 模型

class PublicSlug < ActiveRecord::Base
  extend FriendlyId

  friendly_id :sluggable_name, use: :slugged

  belongs_to :sluggable, polymorphic: true

  private

  # Generate the slug based on the title of the Sluggable class
  def sluggable_name
    sluggable.name
  end
end

Sluggable 问题

module Sluggable
  extend ActiveSupport::Concern

  included do
   has_one :public_slug, dependent: :destroy, as: :sluggable

   delegate :slug, to: :public_slug
  end
end

公司和用户模型。

class User < ActiveRecord::Base
  include Sluggable
end

class Company < ActiveRecord::Base
  include Sluggable
end

我现在可以使用

查询这两个模型

Sluggable.friendly.find(slug).sluggable

重定向可以在你的控制器中处理如下：

def redirect
  @sluggable = Sluggable.friendly.find(params[:search]).sluggable

  redirect_to @sluggable
end

Answer 3

基于@Essn 的回答...

仍然使用 PublicSlug 模型：

# app/models/public_slug.rb
class PublicSlug < ActiveRecord::Base
  extend FriendlyId

  friendly_id :sluggable_name, use: :slugged
  belongs_to :sluggable, polymorphic: true

  validates :slug, presence: true, uniqueness: { case_sensitive: false }

  private
    # Generate the slug based on the title of the Sluggable class
    def sluggable_name
      sluggable.name
    end
end

还有一个值得关注的问题：

# app/models/concerns/sluggable.rb
module Sluggable
  extend ActiveSupport::Concern

  included do
    before_validation :create_public_slug
    has_one :public_slug, dependent: :destroy, as: :sluggable
    delegate :slug, to: :public_slug

    private
      def create_public_slug
        self.public_slug = PublicSlug.new unless public_slug.present?
      end
  end
end

在您要查找的所有模型中包含该关注点：

# app/models/user.rb
class User < ActiveRecord::Base
  include Sluggable
  ...
end

创建迁移：

# db/migrate/...create_public_slugs.rb
class CreatePublicSlugs < ActiveRecord::Migration[5.0]
  def change
    create_table :public_slugs do |t|
      t.references :sluggable, polymorphic: true
      t.string :slug
    end
  end
end

然后您可以通过以下方式查找模型：

# app/controllers/home_controller.rb
class HomeController < ApplicationController
  # /:slug
  def show
    @sluggable = PublicSlug.friendly.find(params[:id]).sluggable
    render @sluggable
  end
end