在 Rails 2.3 中组合选项路由

Question

我正在尝试设置这样的路线：

获取/设置/ 放置/设置/ 获取/设置/照片 PUT /设置/照片

这是我为其设置的路由代码：

#I just do this for code reuse
get = { :method => :get }
put = { :method => :put }

pub.settings '/settings', :controller => :settings, :action => :index, :conditions => get
pub.with_options :controller => :settings, :path_prefix => "/settings", :name_prefix => "settings_" do |settings|
  settings.update '', :action => :update, :conditions => put
  settings.photos '/photos', :action => :photos, :conditions => get
  settings.photos_update '/photos', :action => :photos_update, :conditions => :put
end

这可行，但如果您注意到第一个路由“pub.settings”在 mapped_options 块之外。

如果我是待办事项

pub.with_options :controller => :settings, :path_prefix => "/settings", :name_prefix => "settings_" do |settings|
  settings.root '', :action => :index, :conditions => get
  settings.update '', :action => :update, :conditions => put
  settings.photos '/photos', :action => :photos, :conditions => get
  settings.photos_update '/photos', :action => :photos_update, :conditions => :put
end

然后我会得到（在 rake 路由中）设置的路径将是“settings_root_path”而不是“settings_path”

有谁知道如何将它包含到块中并且仍然将路由函数名称作为“settings_path”？

Answer 1

Rails 有一个:path_prefix、:path_names 和:name_prefix，它们可以帮助控制帮助器的生成方式。 rails routing guide 有几个例子可能会有所帮助。

:path_names 控制在 restful 路由中使用的名称 :path_prefix 设置生成助手时使用的路径。 :name_prefix 将前缀设置为生成的助手的名称。此前缀可以设置为nil 或empty_string。

对于您的情况，我会尝试（尽管未经测试）：

settings.root '', :action => :index, :name_prefix => nil, :conditions => get

Answer 2

 settings.settings  '', :action => :index, :conditions => get, :name_prefix => ''