如何使用带有Rack :: Builder :: map的lotus路由器

Question

有没有办法将地图和(莲花)路由器命名空间一起使用？下面是config.ru我试图作为演示运行的示例.

require 'bundler'
Bundler.require

module Demo

  class Application

    def initialize
      @app = Rack::Builder.new do
        map '/this_works' do
          run  Proc.new {|env| [200, {"Content-Type" => "text/html"}, ["this_works"]]}
        end
        map '/api' do
          run Lotus::Router.new do
            get '/api/', to: ->(env) { [200, {}, ['Welcome to Lotus::Router!']] }
            get '/*', to: ->(env) { [200, {}, ["This is catch all: #{ env['router.params'].inspect }!"]] }
          end
        end
      end
    end

    def call(env)
      @app.call(env)
    end
  end  
end

run Demo::Application.new

Answer 1

您的问题是由于do..end方法调用的优先级.在你的代码中的部分

run Lotus::Router.new do
  get '/api/', to: ->(env) { [200, {}, ['Welcome to Lotus::Router!']] }
  get '/*', to: ->(env) { [200, {}, ["This is catch all: #{ env['router.params'].inspect }!"]] }
end

被Ruby解析为

run(Lotus::Router.new) do
  get '/api/', to: ->(env) { [200, {}, ['Welcome to Lotus::Router!']] }
  get '/*', to: ->(env) { [200, {}, ["This is catch all: #{ env['router.params'].inspect }!"]] }
end

换句话说,块传递给run,而不是Lotus::Router.new按照您的意图传递,并run简单地忽略块.

要修复它,您需要确保该块与路由器的构造函数相关联,而不是调用run.有几种方法可以做到这一点.您可以使用{...}而不是do...end,因为它具有更高的优先级:

run Lotus::Router.new {
  #...
}

另一种方法是将路由器分配给局部变量,并将其用作以下参数run:

router = Lotus::Router.new do
  #...
end
run router