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Home/ Questions/Q 7735513
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T07:31:16+00:00 2026-06-01T07:31:16+00:00

There are two ways to write a polymorphic migration in Rails. Generally, I’ve done

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There are two ways to write a polymorphic migration in Rails. Generally, I’ve done this:

class CreateFeatures < ActiveRecord::Migration
  def change
    create_table :features do |t|
      t.integer  :featureable_id
      t.string   :featurable_type

      t.timestamps
    end
  end
end

However, we can also do this:

class CreateFeatures < ActiveRecord::Migration
  def change
    create_table :features do |t|
      t.references  :featureable, :polymorphic => true

      t.timestamps
    end
  end
end

The two are, for all practical purposes, identical. My question: Is one better than another? Is one better for future maintainability?

This would likely be an issue only if one of two things changed:

  1. The polymorphic abstraction version (Version #2) goes away or the syntax changes
  2. The method of working a polymorphic relationship (using id and type) changes- unlikely

Just wondering if there’s a preference, or if it’s “Meh, doesn’t really matter either way”

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  1. Editorial Team
    Editorial Team
    2026-06-01T07:31:18+00:00Added an answer on June 1, 2026 at 7:31 am

    For an all rails app, where you are generating all tables via migrations, there is no difference functionally.

    Here is the code for references:

    def references(*args)
      options = args.extract_options!
      polymorphic = options.delete(:polymorphic)
      args.each do |col|
        @base.add_column(@table_name, "#{col}_id", :integer, options)
        @base.add_column(@table_name, "#{col}_type", :string, polymorphic.is_a?(Hash) ? polymorphic : options) unless polymorphic.nil?
      end
    end
    

    This is all well and good but if your foreign keys on the referenced table are not _id, method one is the only choice.

    references just saves you one line of code…

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