How to get last N records with activerecord?

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

This is the Rails 3 way

SomeModel.last(5) # last 5 records in ascending order

SomeModel.last(5).reverse # last 5 records in descending order

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

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

For Rails 5 (and likely Rails 4)

Bad:

Something.last(5)

because:

Something.last(5).class
=> Array

so:

Something.last(50000).count

will likely blow up your memory or take forever.

Good approach:

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

because:

Something.limit(5).order('id desc').class
=> Image::ActiveRecord_Relation

Something.limit(5).order('id desc').to_sql
=> "SELECT  \"somethings\".* FROM \"somethings\" ORDER BY id desc LIMIT 5"

The latter is an unevaluated scope. You can chain it, or convert it to an array via .to_a. So:

Something.limit(50000).order('id desc').count

... takes a second.

For Rails 4 and above version:

You can try something like this If you want first oldest entry

YourModel.order(id: :asc).limit(5).each do |d|

---

You can try something like this if you want last latest entries..

YourModel.order(id: :desc).limit(5).each do |d|

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".

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

Just try:

Model.order("field_for_sort desc").limit(5)

In my rails (rails 4.2) project, I use

Model.last(10) # get the last 10 record order by id

and it works.

we can use Model.last(5) or Model.limit(5).order(id: :desc) in rails 5.2

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.

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.

If you have a default scope in your model that specifies an ascending order in Rails 3 you'll need to use reorder rather than order as specified by Arthur Neves above:

Something.limit(5).reorder('id desc')

or

Something.reorder('id desc').limit(5)

A simple answer would be:

Model.limit(5).order(id: :desc)

There is a problem with this solution, as id can't be the sole determiner of when a record was created in the time.

A more reliable solution would be:

Model.order(created_at: :desc).limit(5)

As others have pointed out, one can also use Model.last(5). The only gotcha with this is that it returns Array, and not Model::ActiveRecord_Relation.

Add an :order parameter to the query