ActiveRecord unknown column-Collection of common programming errors

I ran into a little problem I have a has_many through relationship here is the code for the models

class User < ActiveRecord::Base
  has_many :friendships
  has_many :followings, :through => :friendships, :foreign_key => "followed_id"
end

class Friendship < ActiveRecord::Base
  belongs_to :user 
  belongs_to :following, :class_name => "User", :foreign_key => "followed_id"
end

now at the console I can type u = User.first and then u.friendships.first.following this gives me the first user u is following, but when I type u.friendships.last.following I get this error

the SELECT statement from u.friendships.first.following

Friendship Load (0.3ms)  SELECT `friendships`.* FROM `friendships` WHERE `friendships`.`user_id` = 208 LIMIT 1
User Load (0.2ms)  SELECT `users`.* FROM `users` WHERE `users`.`id` = 209 LIMIT 1

and the SELECT statement from u.friendships.last.following

Friendship Load (0.3ms)  SELECT `friendships`.* FROM `friendships` WHERE `friendships`.`user_id` = 208 ORDER BY `friendships`.`` DESC LIMIT 1

ActiveRecord::StatementInvalid: Mysql2::Error: Unknown column 'friendships.' in 'order
clause': SELECT  `friendships`.* FROM `friendships`  WHERE `friendships`.`user_id` = 208
ORDER BY `friendships`.`` DESC LIMIT 1

if I then run u.friendships and then u.friendships.last.following again, I don’t get the error anymore, why is that?

1. Heres my sql output for friendships, straight from your code on Rails 3.2.9 / postgresql:

   # u.friendships.first.following
   Friendship Load (0.9ms)  SELECT "friendships".* FROM "friendships" WHERE "friendships"."user_id" = 1 LIMIT 1
   
   # u.friendships.first.following
   Friendship Load (1.3ms)  SELECT "friendships".* FROM "friendships" WHERE "friendships"."user_id" = 1 ORDER BY "friendships"."id" DESC LIMIT 1
   

   So for some reason for me, id is getting picked up automatically in ORDER BY "friendships"."id" and it works. Maybe your problem has something to do with your DB?

   #Statements used to create the db for reproducing this problem
   CREATE TABLE users (id SERIAL PRIMARY KEY)
   CREATE TABLE friendships (
       id            SERIAL PRIMARY KEY,
       user_id       integer
       followed_id   integer 
   );
   

Originally posted 2013-11-10 00:11:12.