Padrino支持嵌套路由的想法。以下是文档中的一个示例:
SimpleApp.controllers :product, :parent => :user do
get :index do
# "/user/#{params[:user_id]}/product"
end
get :show, :with => :id do
# "/user/#{params[:user_id]}/product/show/#{params[:id]}"
end
end
但是,我想要的是能够拥有以下映射:
GET /users # '/' in :users controller
GET /users/:id # '/:id' in :users controller
GET /users/:user_id/tweets # '/' in :tweets controller
GET /users/:user_id/tweets/:id # '/:id' in :tweets controller
GET /tweets # '/' in :tweets controller, too
GET /tweets/:id # '/:id' in :tweets controller, too
这可能吗?
答案 0 :(得分:0)
SimpleApp.controllers :tweets, :parent => :user do
get :index do
# "/user/#{params[:user_id]}/tweets"
end
get :index, :with => :id do
# "/user/#{params[:user_id]}/tweets/#{params[:id]}"
end
end
答案 1 :(得分:0)
您是否尝试过:optional =>您的路线定义(第1行)是真的吗?