How to get last N records with activerecord?

Question

With :limit in query, I will get first N records. What is the easiest way to get last N records?

Answer 1

If you need to set some ordering on results then use:

Model.order('name desc').limit(n) # n= number

if you do not need any ordering, and just need records saved in the table then use:

Model.last(n) # n= any number

Answer 2

Updated Answer (2020)

You can get last N records simply by using last method:

Record.last(N)

Example:

User.last(5)

Returns 5 users in descending order by their id.

Deprecated (Old Answer)

An active record query like this I think would get you what you want ('Something' is the model name):

Something.find(:all, :order => "id desc", :limit => 5).reverse

edit: As noted in the comments, another way:

result = Something.find(:all, :order => "id desc", :limit => 5)

while !result.empty?
        puts result.pop
end

Answer 3

new way to do it in rails 3.1 is SomeModel.limit(5).order('id desc')

Answer 4

Let's say N = 5 and your model is Message, you can do something like this:

Message.order(id: :asc).from(Message.all.order(id: :desc).limit(5), :messages)

Look at the sql:

SELECT "messages".* FROM (
  SELECT  "messages".* FROM "messages"  ORDER BY "messages"."created_at" DESC LIMIT 5
) messages  ORDER BY "messages"."created_at" ASC

The key is the subselect. First we need to define what are the last messages we want and then we have to order them in ascending order.

Answer 5

Solution is here:

SomeModel.last(5).reverse

Since rails is lazy, it will eventually hit the database with SQL like: "SELECT table.* FROM table ORDER BY table.id DESC LIMIT 5".

Answer 6

I find that this query is better/faster for using the "pluck" method, which I love:

Challenge.limit(5).order('id desc')

This gives an ActiveRecord as the output; so you can use .pluck on it like this:

Challenge.limit(5).order('id desc').pluck(:id)

which quickly gives the ids as an array while using optimal SQL code.