How to get last N records with activerecord? - ruby-on-rails

Question

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

Answer 1

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

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

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.

Answer 5

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|

Answer 6

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 7

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 8

In my rails (rails 4.2) project, I use
Model.last(10) # get the last 10 record order by id
and it works.

Answer 9

Just try:
Model.order("field_for_sort desc").limit(5)

Answer 10

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

Answer 11

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.

Answer 12

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 13

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)

Answer 14

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.

Answer 15

Add an :order parameter to the query