Simple Rails routing situation

Question

I'm creating a Ruby on Rails app that consists of stories. Each story has multiple pages.

How can I set up routes.rb so that I can have URLs like this:

http://mysite.com/[story id]/[page id]

Like:

http://mysite.com/29/46

Currently I'm using this sort of setup:

http://mysite.com/stories/29/pages/46

Using:

ActionController::Routing::Routes.draw do |map|

  map.resources :stories, :has_many => :pages
  map.resources :pages

  map.root :controller => "stories", :action => "index"

  map.connect ':controller/:action/:id'
  map.connect ':controller/:action/:id.:format'

end

Thanks in advance. I'm a newbie to Rails and routing seems a bit complicated to me right now.

Answer 1

Here's a good resource.

map.connect 'stories/:story_id/:page_num', :controller => "stories", :action => "index", :page_id => nil

In your controller:

def index
  story = Story.find(params[:story_id])
  @page = story.pages[params[:page_num]]
end

UPDATE: Just noticed you don't want 'stories' to appear. Not sure if dropping that from the map.connect will work... try it out.

map.connect '/:story_id/:page_num', ...

Answer 2

Terry Lorber's answer, above, works very well, provided you don't forget the slash at the beginning of his revised solution, like so:

map.connect '/:story_id/:page_id', :controller => "pages", :action => "index", :page_id => nil

But I don't know where to go from here! How do I get the controller to go to the correct page?

Answer 3

Given the route of

map.connect '/:story_id/:id', :controller => 'pages', :action => 'show'

You could do this in your controller

def show
  @story = Story.find(:story_id)
  @page = @story.pages.find(:id)
end

Now you can get the page using the story id.

PS: map.connect ':story_id/:id' should work.

Answer 4

I am not sure why no one has suggested just using a named route. Something like

map.story '/:story_id/:page_id', :controller => 'stories', :action => 'show'

then you could just call it with

story_path(story.id,params[:page})

or something similar.