Rails destroy all but newest n records

2020-06-30 11:50发布

问题:

How do I destroy all but the newest n records using Rails' ActiveRecord?

I can get the newest n records using order and limit but how do I destroy the inverse?

回答1:

Either of these methods would do it:

# Fetch your latest N records
newest_n_records = Foo.find(:all, :order => 'created_at DESC', :limit => n)

# Then do:
Foo.destroy_all(['id NOT IN (?)', newest_n_records.collect(&:id)])

# Or:
Foo.destroy_all('created_at < ?', newest_n_records.last.created_at)


回答2:

I have two methods of doing this, assuming n = 5:

Foo.order('id desc').offset(5).destroy_all

This sorts records with latest first, and destroys everything past the 5th records. Or

Foo.destroy_all(['id <= ?', Foo.order('id desc').limit(1).offset(5).first.id])

This finds the 6th latest record id and deletes all records with id <= 6th latest record id.

Also, you might want to look at this SO question.



回答3:

Foo.destroy_all(['id NOT IN (?)', Foo.last(1000).collect(&:id)])


回答4:

Person.destroy_all("last_login < '2004-04-04'")

This will destroy all persons who meet the condition. So all you need is inverted conditions and destroy_all



回答5:

Previous answers use find or last require the creation of ActiveModel, which take extra computation time.

I think using pluck is better, since it only creates an Array of ids.

ids = Foo.limit(n).order('id DESC').pluck(:id)
Foo.where('id NOT IN (?)', ids).destroy_all


回答6:

[Rails 5 / ActiveRecord::Relation]

destroy_all no longer takes parameters... Actually, the ActiveRecord::Relation never allowed parameters I don't think... Anyway, you should just put the condition before it, but use destroy_all after the query, like this:

Person.destroy_all("last_login < '2004-04-04'")
Person.destroy_all(status: "inactive")
Person.where(age: 0..18).destroy_all