Problem with has_many :through Association

Hi, I've got a problem with a has_many :through association, i cant
call the u1.UsersProfileAttributes.find_by_ProfileAttribute_name
("icq") method, rails means that this method doesn't exist. The method
u1.UsersProfileAttributes.find_by_ProfileAttribute_id(3) works
correctly. u1 is a user object. I don't know whats the problem,
because my associations seems to be okay.

Have a look:

class ProfileAttribute < ActiveRecord::Base
  has_many :UsersProfileAttributes
  has_many :users, :through => :UsersProfileAttributes
end
class User < ActiveRecord::Base
  has_many :UsersProfileAttributes
  has_many :ProfileAttributes, :through => :UsersProfileAttributes
end
class UsersProfileAttribute < ActiveRecord::Base
  belongs_to :user
  belongs_to :ProfileAttribute
end

My Migration file:

    class CreateProfileAttributes < ActiveRecord::Migration
      def self.up
        create_table :profile_attributes do |t|
          t.string :name
          t.integer :profile_group_id
          t.timestamps
        end
    create_table :users_profile_attributes do |t|
      t.integer :user_id
      t.integer :ProfileAttribute_id
      t.string :value
    end
  end
  def self.down
    drop_table :profile_attributes
    drop_table :users_profile_attributes
  end
end

Hi, I've got a problem with a has_many :through association, i cant
call the u1.UsersProfileAttributes.find_by_ProfileAttribute_name
("icq") method, rails means that this method doesn't exist. The method
u1.UsersProfileAttributes.find_by_ProfileAttribute_id(3) works
correctly. u1 is a user object. I don't know whats the problem,
because my associations seems to be okay.

has_many :UserProfileAttributes etc... should really be
has_many :user_profile_attributes. Looks like you're getting away with
it for now but you're probably storing up trouble for now.

u1.UsersProfileAttributes.find_by_ProfileAttribute_name doesn't work
because the users_profile_attributes tables doesn't have a name column
( but it does have a ProfileAttribute_id so your second example does
work).

Fred