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Nesta CMS and Rails3 integration: adding blog to an exiting rails 3 application

开发者 https://www.devze.com 2023-03-30 19:27 出处:网络
I\'m adding nesta (0.9.8) cms, to an existing Rails 3.0.10 application. I get the blog up and running but not the layout/stylesheets.

I'm adding nesta (0.9.8) cms, to an existing Rails 3.0.10 application. I get the blog up and running but not the layout/stylesheets.

What i did until now is : 1. inside rails app main root, add gem 'nesta', gem 'sass' and run 'bundle'

2. run "nesta new nesta-blog" 3. edit config.ru like following :

require ::File.expand_path('../config/environment',  __FILE__)
map "/" do
 run MyRails3App::Application
end

require 'nesta/env'
require 'nesta/app'

Nesta::App.root = ::File.expand_path('./nesta-blog', ::File.dirname(__FILE__))
map "/blog" do
 run Nesta::App
end

4. edit config/routes.rb like following :

require 'nesta/env'
require 'nesta/app'

Rails3MongoidOmniauthSimple::Application.routes.draw do

 mount Nesta::App.new => "/blog"
 root :to => "home#index"
...

5. cd nesta-blog 6. run n开发者_StackOverflowesta demo:content

Now, if you run rails s from your ~/main-rails-app, going to http://localhost:3000/blog you will see the the demo nesta site but without his default layout/stylesheets, while if you run shotgun config.ru from inside ~/main-rails-app/nesta-blog, going to http://localhost:9393/ everything shows up correctly.

Any suggestion?

Thanks in advance Luca G. Soave


I've not got this to the level of plug-n-play that I'd like yet, but I'm running Nesta on my Rails 3.0 sites by adding this to config/routes.rb:

mount Nesta::App, :at => '/'
match '/css/*style.css' => Nesta::App
match '/attachments/*file' => Nesta::App

I haven't looked into a cleaner way of doing this yet (i.e. avoiding having to specify css and attachments routes as well).

I created my Nesta app in a directory located at "#{Rails.root}/nesta". I also needed an in config/initializers/nesta.rb:

require "nesta/env"
Nesta::Env.root = ::File.expand_path("../../nesta",
                                     File.dirname(__FILE__))

I quite like the way you've done it too.

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