我有一个共同的基本路径;说：get / base我需要执行基本身份验证并在该路径下为所有子调用工作.说：get / base / foo和get / base / bar.
看http://www.sinatrarb.com/intro.html#Helpers表明我应该能够使用助手来做到这一点.我正在查看传递帮助程序并在文档中触发新路由时使用调用.但是,我读到的另一个建议是使用正则表达式IE％r {/ base /？:( path)？}或其他一些动态路由.那么怎么样：
def '/base'
# do some funky basic auth stuff here
# to work with all request to this common
# base path?
pass
end
def %r{/base/?(path)?} do |path|
case path
when 'foo'
# do something.
when 'bar'
# do something else.
end
# some kind of redirection or template rendering here:
erb :template
end
有没有人以前处理过这种事情？我想保持干燥.当然,我不确定给定的例子在保留params方面是最好的.
解决方法
有几种方法可以做到这一点(这些不是任何顺序,它们都很好)：
具有前块和助手的命名空间
http://rubydoc.info/gems/sinatra-contrib/1.3.2/Sinatra/Namespace
require 'sinatra/namespace'
class App < Sinatra::Base
register Sinatra::Namespace
namespace "/base" do
helpers do # this helper is now namespaced too
authenticate!
# funky auth stuff
# but no need for `pass`
end
end
before do
authenticate!
end
get "/anything" do
end
get "/you" do
end
get "/like/here" do
end
end
带有条件的命名空间
require 'sinatra/namespace'
class App < Sinatra::Base
register Sinatra::Namespace
set(:auth) do |*roles| # <- notice the splat here
condition do
unless logged_in? && roles.any? {|role| current_user.in_role? role }
redirect "/login/",303
end
end
end
namespace "/base",:auth => [:user] do
# routes…
end
namespace "/admin",:auth => [:admin] do
# routes…
end
助手和前过滤器
helpers do
authenticate!
# funky auth stuff
# but no need for `pass`
end
end
before '/base/*' do
authenticate!
end
映射的应用
class MyBase < Sinatra::Base
helpers do
authenticate!
# funky auth stuff
# but no need for `pass`
end
end
before do
authenticate!
end
get "/" do
end
get "/another" do
end
end
# in rackup file
map "/" do
run App1
end
map "/base" do
# every route in MyBase will now be accessed by prepending "/base"
# e.g. "/base/" and "/base/another"
run MyBase
end
#…
我不确定是否需要使用案例陈述来干预路线.如果路线各自做了不同的事情,那么我只是单独写出来,因为它更加清晰,你正在重复Sinatra在匹配路线中所做的工作.