如何创建这条路线？

Question

我有以下路线：

resources :articles, except: [:index] do
   collection do
      get 'author/:id/articles' => 'articles#index', as: 'index_articles'
   end
end

这导致：

GET    /api/v1/articles/:id/author/:id/articles(.:format)

如何将其转换为以下内容？：

GET    /api/v1/author/:id/articles(.:format)  

Answer 1

# config/routes.rb
resources :articles, except: [:index]
resources :authors, path: "author", only: [] do
   resources :articles, only: :index, on: :member #-> url.com/author/:id/articles
end

Answer 2

写这个

get 'author/:id/articles' => 'articles#index', as: 'index_articles'

在resources :articles区块之外。