使用RegExp的Sinatra路由参数

Question

我正在尝试创建一个小的sinatra示例,它将打印一个月,日或年的日历视图,具体取决于URL.这样

localhost:4567/calendar/2012将打印年视图日历

localhost:4567/calendar/2012/9将打印一个月的日历

localhost:4567/calendar/2012/9/15将打印日视图日历

我有这个工作:

require 'sinatra'

get '/calendar/:year/:month/:day/?' do
    "printing daily calendar for #{params[:year]}/#{params[:month]}/#{params[:day]}"
end

get '/calendar/:year/:month/?' do
    "printing monthly calendar for #{params[:year]}/#{params[:month]}"
end

get '/calendar/:year/?' do
    "printing yearly calendar for #{params[:year]}"
end

我的问题是,我是否可以使用某种RegEx进一步优化这些路由,以便我可以说如果:month部分在1到12之间并且:day部分在1到31之间,则仅计算url？

Answer 1

如果您使用的是ruby 1.9+,则可以使用这样的命名捕获:

get %r{/(?<year>\d{4})/(?<month>\d{2})/(?<day>\d{2})/?} do
 "printing daily calendar for #{params[:year]}/#{params[:month]}/#{params[:day]}"
end

get %r{/(?<year>\d{4})/(?<month>\d{2})/?} do
  "printing monthly calendar for #{params[:year]}/#{params[:month]}"
end

get %r{/(?<year>\d{4})/?} do
  "printing monthly calendar for #{params[:year]}"
end