如何访问 Rails 3 应用程序对象的名称？

Question

我需要从 Rails 3 引擎中知道程序员应用程序的名称。

Answer 1

Rails.application.class.parent_name

Answer 2

The original poster asked for the name of the rails app , not for the class name. Those are two different things. eg the spelling of the rails app can be different from what Rails expects, eg 'test-app' instead of 'test_app'.

It is difficult to get to the name of the Rails App spelled the way as it was checked-in , because that name is not preserved. Only the options for the session_store seem to contain the original string slightly modified.

The best way to get to the name of the Rails application is:

This will work even if your app directory was renamed, or sym-linked!

Rails.application.config.session_options[:key].sub(/^_/,'').sub(/_session/,'')
=> "test-app"

Why? Because the author of the app could have spelled it's name differently than Rails expects... eg with '-' characters instead of '_'; eg "test-app". From the class name you can't guarantee to get to the correct spelling.

Using this info, you could do this:

class << Rails.application
  def name
    Rails.application.config.session_options[:key].sub(/^_/,'').sub(/_session/,'')
  end
end

Rails.application.name
=> 'test-app'

or just add this to your ./config/environment.rb :

APP_VERSION = '1.0.0'
APP_NAME = Rails.application.config.session_options[:key].sub(/^_/,'').sub(/_session/,'')

that makes these constants available on the top-level of the app.

---

Close, but no cigar:

This is almost correct, but it will still fail if the application directory gets renamed (eg during deployment to '"20121001_110512" or "latest"... then the following would break:

File.basename(Rails.root.to_s)
=> "test-app"

---

with the following two approaches you can't get to the correct spelling.. you could only guess a name:

This is sub-optimal, and can give incorrectly spelled results:

You can get to a derivative of the application name like this:

Rails.application.engine_name.gsub(/_application/,'')
=> "test_app"

But please note that this is not fool-proof, because somebody could have named the app "test-app" and you will see the result above, not the correct name with '-'.

Same thing is true if you derive it like this:

Rails.application.class.parent_name
=> "TestApp"

this is the class name, but you can't be sure how to get to the name the way the author spelled it.

Answer 3

In Rails 3, the application that is generated is given a module namespace matching the application name. So if your application was called "Twitter", the class of Rails.application is "Twitter::Application".

This means that you could split the string of the class name of your Rails application to get an application name like this:

Rails.application.class.to_s.split("::").first

In our example, the resulting string would be "Twitter".

Answer 4

Rails.application返回您的MyApp::Application类。

Answer 5

For a more "humanized" version (based on the first reply you can find here) consider the following method: it will return Test App for your TestApp application.

def app_name
  Rails.application.class.parent_name
    .underscore
    .humanize
    .split
    .map(&:capitalize)
    .join(' ')
end

Answer 6

For Rails 6 and beyond

Rails.application.class.module_parent.name

More context:

Module#parent has been renamed to module_parent . parent is deprecated and will be removed in Rails 6.1.

Answer 7

Using Scott's answer, I put the following in my app/controllers/application_controller.rb file

class ApplicationController < ActionController::Base
  protect_from_forgery
  helper_method :app_name
  private
    def app_name
      Rails.application.class.to_s.split("::").first
    end

Then in my templates I can just do #{app_name} wherever it's needed, which makes it much less of a hassle if you need to rename your app (which I had to do this morning).

Answer 8

You can find this in the config.ru file:

run Dinosaur::Application

or in the Rakefile:

Dinosaur::Application.load_tasks

The name prior to the "::" is the app name