How do you find the namespace/module name programmatically in Ruby on Rails?

Question

How do I find the name of the namespace or module 'Foo' in the filter below?

class ApplicationController < ActionController::Base
  def get_module_name
    @module_name = ???
  end
end


class Foo::BarController < ApplicationController
  before_filter :get_module_name
end

Bridge

Answer 1

None of these solutions consider a constant with multiple parent modules. For instance:

A::B::C

As of Rails 3.2.x you can simply:

"A::B::C".deconstantize #=> "A::B"

As of Rails 3.1.x you can:

constant_name = "A::B::C"
constant_name.gsub( "::#{constant_name.demodulize}", '' )

This is because #demodulize is the opposite of #deconstantize:

"A::B::C".demodulize #=> "C"

If you really need to do this manually, try this:

constant_name = "A::B::C"
constant_name.split( '::' )[0,constant_name.split( '::' ).length-1]

Ryanryann

This question was written before Rails 3, and modules weren't really supported. Updated accepted answer to reflect more current versions! – Bridge

demodulize was the key answer – Tamworth

Answer 2

For the simple case, You can use :

self.class.parent

Rosewood

Is that available only in recent Rubies ? Can't believe there are so many answers with "dark magic" and none referenced this method until 2015 – Dryer

Can't find it in the doc... I can't even reproduce :) I prefere this answer now : #14779316 : class C;end; C.new.class.superclass -> Object – Rosewood

I think my solution worked because of rails : rubydoc.info/docs/rails/4.1.7/Module#parent-instance_method – Rosewood

Answer 3

For Rails 6.1

self.class.module_parent

---

Hettomei answer works fine up to Rails 6.0

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

Forerun

module_parent has been around since Rails 6.0 – Roehm

This is the type of clean approach I was looking for. Thanks! – Rothberg

Answer 4

This should do it:

  def get_module_name
    @module_name = self.class.to_s.split("::").first
  end

Bondholder

or u can also use self.class.name.demodulize for that purpose. – Cruse

no you can't, actually demodulize does return class name without namespaces – Document

As Jason Harrelson pointed out, this does not take into consideration multiple parent modules. – Bilateral

Answer 5

This would work if the controller did have a module name, but would return the controller name if it did not.

class ApplicationController < ActionController::Base
  def get_module_name
    @module_name = self.class.name.split("::").first
  end
end

However, if we change this up a bit to:

class ApplicatioNController < ActionController::Base
  def get_module_name
    my_class_name = self.class.name
    if my_class_name.index("::").nil? then
      @module_name = nil
    else
      @module_name = my_class_name.split("::").first
    end
  end
end

You can determine if the class has a module name or not and return something else other than the class name that you can test for.

Bridge

Answer 6

No one has mentioned using rpartition?

const_name = 'A::B::C'
namespace, _sep, module_name = const_name.rpartition('::')
# or if you just need the namespace
namespace = const_name.rpartition('::').first

Tympany

Answer 7

I know this is an old thread, but I just came across the need to have separate navigation depending on the namespace of the controller. The solution I came up with was this in my application layout:

<%= render "#{controller.class.name[/^(\w*)::\w*$/, 1].try(:downcase)}/nav" %>

Which looks a bit complicated but basically does the following - it takes the controller class name, which would be for example "People" for a non-namespaced controller, and "Admin::Users" for a namespaced one. Using the [] string method with a regular expression that returns anything before two colons, or nil if there's nothing. It then changes that to lower case (the "try" is there in case there is no namespace and nil is returned). This then leaves us with either the namespace or nil. Then it simply renders the partial with or without the namespace, for example no namespace:

app/views/_nav.html.erb

or in the admin namespace:

app/views/admin/_nav.html.erb

Of course these partials have to exist for each namespace otherwise an error occurs. Now the navigation for each namespace will appear for every controller without having to change any controller or view.

Laryngology

Answer 8

my_class.name.underscore.split('/').slice(0..-2)

or

my_class.name.split('::').slice(0..-2)

Jez

Answer 9

With many sub-modules:

module ApplicationHelper
  def namespace
    controller.class.name.gsub(/(::)?\w+Controller$/, '')
  end
end

Example: Foo::Bar::BazController => Foo::Bar

Peerage

Answer 10

I don't think there is a cleaner way, and I've seen this somewhere else

class ApplicationController < ActionController::Base
  def get_module_name
    @module_name = self.class.name.split("::").first
  end
end

Melton

Answer 11

I recommend gsub instead of split. It's more effective that split given that you don't need any other module name.

class ApplicationController < ActionController::Base
  def get_module_name
    @module_name = self.class.to_s.gsub(/::.*/, '')
  end
end

Ulmaceous