Parameters 友好的\u Id:Id是slug，但url仍然显示对象Id属性（？Id=）

我试图在rails 4应用程序中实现友好的_Id gem，但遇到了一个障碍

gem正在运行，但是url仍然不太好。看起来是这样的：

resources :projects do
  member do
    get :featured
  end
end

gem正在工作，因为id参数现在是slug，但是我不知道如何删除“？id=”并用斜杠替换它？可能吗？我漏了一步吗

控制器：

  def featured
    @project = Project.friendly.find(params[:id])
  end

  extend FriendlyId
  friendly_id :title, use: :slugged

     <% @projects.each do |project| %>

        <%= link_to project_featured_path(:id => project.slug) do %>

            <%= project.title %>

        <% end %>

    <% end %>

get "project/featured"

match 'project' => 'project#index', :as => :project, via: [:get, :post]

resources :projects do
  member do
    get :featured
  end
end

match ':controller(/:action(/:id))(.:format)', via: [:get, :post]   

型号：

  def featured
    @project = Project.friendly.find(params[:id])
  end

  extend FriendlyId
  friendly_id :title, use: :slugged

     <% @projects.each do |project| %>

        <%= link_to project_featured_path(:id => project.slug) do %>

            <%= project.title %>

        <% end %>

    <% end %>

get "project/featured"

match 'project' => 'project#index', :as => :project, via: [:get, :post]

resources :projects do
  member do
    get :featured
  end
end

match ':controller(/:action(/:id))(.:format)', via: [:get, :post]   

查看：

  def featured
    @project = Project.friendly.find(params[:id])
  end

  extend FriendlyId
  friendly_id :title, use: :slugged

     <% @projects.each do |project| %>

        <%= link_to project_featured_path(:id => project.slug) do %>

            <%= project.title %>

        <% end %>

    <% end %>

get "project/featured"

match 'project' => 'project#index', :as => :project, via: [:get, :post]

resources :projects do
  member do
    get :featured
  end
end

match ':controller(/:action(/:id))(.:format)', via: [:get, :post]   

我还引用了我以前在rails 3中运行Friendly_Id的应用程序。它似乎没有这个问题。。。不记得我是不是错过了一步

任何帮助都将不胜感激。干杯

更新：

  def featured
    @project = Project.friendly.find(params[:id])
  end

  extend FriendlyId
  friendly_id :title, use: :slugged

     <% @projects.each do |project| %>

        <%= link_to project_featured_path(:id => project.slug) do %>

            <%= project.title %>

        <% end %>

    <% end %>

get "project/featured"

match 'project' => 'project#index', :as => :project, via: [:get, :post]

resources :projects do
  member do
    get :featured
  end
end

match ':controller(/:action(/:id))(.:format)', via: [:get, :post]   

替换后：

project\u特色路径（：id=>project.slug）

答复如下：

project\u特色路径（project）

我的url看起来好一点，但仍然不起作用。看起来是这样的：

resources :projects do
  member do
    get :featured
  end
end

有一个句号而不是斜杠。在测试中，我发现如果我只是用斜杠替换了点，那么实际上就生成了正确的页面

---

这只是我的链接格式有问题吗？

答案取决于您的路线，因此发布它们可能也很方便，但如果它们看起来像这样：

resources :projects do
  member do
    get :featured
  end
end

我猜您想要的语法是：

project_featured_path(project)

我签出了friendly_id，friendly_id似乎覆盖了活动记录“to_param”方法。rails路径/url帮助程序调用它来从AR对象构造url，因此将使用url slug构造url

另请参阅中的finder方法，其中友好ID似乎覆盖了“find”方法。它首先寻找一个友好的id，如果失败，它会返回到传统的id来查找

---

我认为这应该会有帮助，或者希望是一个正确方向的提示。

谢谢您的快速回复！