1

用户可以有多个消息:

class User < ActiveRecord::Base
  has_many :messages, :foreign_key => "publisher_id"
end

要获得最新的消息时间,我会这样做:

class User < ActiveRecord::Base
  def last_message_time
    last_message = messages.max{ |m1, m2| m1.created_at <=> m2.created_at }
    last_message ? last_message.created_at : nil
  end
end

出于某种原因,我确信有更好的方法来做到这一点。

你会怎么做?

4

2 回答 2

6

这是一个简短的版本:

# Ruby < 2.3.0
def last_message_time
  messages.order(:created_at).last.try(:created_at)
end

# Ruby >= 2.3.0
def last_message_time
  messages.order(:created_at).last&.created_at
end

更新(谢谢@mnort9): 为 Ruby > 2.3.0 更新

于 2011-08-09T13:11:56.913 回答
4

至少,您可以让数据库为您进行排序。

class User < ActiveRecord::Base
  def last_message_time
    last_message = messages.find(:last, :order => :created_at)
    last_message ? last_message.created_at : nil
  end
end
于 2011-08-09T13:09:27.953 回答