如何从URL中轨删除动作和控制器的link_to

Question

这里是我的link_to

<%= link_to sub_category.name, controller: :posts, action: :product, id: "#{sub_category.slug}-#{sub_category.id}" %> 

这是指向URL

http://localhost:3000/posts/product/fifith-category-s-sub-category-2 

我需要的网址如下

http://localhost:3000/fifith-category-s-sub-category-2 

我该怎么做。

我route.rb

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

Answer 1

什么@MarekLipka建议是正确的，但你的定义这样的路线将占用您应用程序中的所有单层命名空间，即“/ anything”将默认路由到posts#product。

我建议使用某种形式的标识符来确定哪些路由应该转到posts#product。什么会为你工作取决于你为什么要这样做。选项夫妇是：

使用短命名空间：

scope '/pp' do 
    get ':id', to: 'posts#product 
end 
# "/pp/:id" routes to 'posts/product' 
# pp is a random short name I picked, it could be anything 

# link 
<%= link_to sub_category.name, "pp/#{sub_category.slug}-#{sub_category.id}" %> 

使用约束：

get ':id', to: 'posts#product`, constraints: { :id => /sub\-category/ } 
# only id's with 'sub-cateogry' route to 'posts/product' 

# link (assuming that sub_category.slug has 'sub-category' words in it) 
<%= link_to sub_category.name, "#{sub_category.slug}-#{sub_category.id}" %> 

Answer 2

如果你想路径/:id符合您posts#product，你应该在你的路线是这样的：

resources :posts 
match ':id', to: 'posts#product`, via: :get 
match ':controller(/:action(/:id))(.:format)', via: [:get, :post]