How to access the object name of a Rails 3 application?

Question

I need to know the name of the Rails 3 programmer application.

Answer 1

Rails.application.class.parent_name 

Answer 2

The original poster asked for the application name rails , not the class name. These are two different things. for example, the spelling of a rails application may be different from what Rails expects, for example. 'test-app' instead of 'test_app'.

It is difficult to get to the name of the Rails application, as noted in , because this name is not saved. Only the parameters for session_store seem to contain the original string slightly modified.

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

This will work even if your application directory has been renamed or associated with a symbol!

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

Why? Because the author of the application could write the name differently than Rails expects ... for example. with the characters '-' instead of '_'; for example, "Test Application." From the name of the class, you cannot guarantee the correct spelling.

Using this information, you can 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/,'') 

making these constants available at the top level of the application.

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Close but no cigar:

This is almost correct, but it will still fail if the application directory is renamed (for example, during deployment to "20121001_110512" or "last" ... then the following broke:

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

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with the following two approaches you cannot get the correct spelling .. you could only guess the name:

This is not optimal and may produce incorrectly generated results:

You can get the derivative on behalf of the application, for example:

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

But keep in mind that this is not perfect, because someone could call the application "test-app" and you will see the result above, and not the correct name with "-".

The same is true if you print it like this:

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

this is the name of the class, but you cannot be sure how to get to the name as the author wrote it.

Answer 3

In Rails 3, the application namespace is set to the module namespace corresponding to the application name. Therefore, if your application was named "Twitter", the Rails.application class will be "Twitter :: Application".

This means that you can 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 will be "Twitter."

Answer 4

Rails.application returns the MyApp::Application class.

Answer 5

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

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

Answer 6

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 is needed, which makes it much less complicated if you need to rename your application (which I should have done this morning).

Answer 7

You can find this in the config.ru file:

 run Dinosaur::Application 

or in the Rakefile:

 Dinosaur::Application.load_tasks 

The name before "::" is the name of the application