2

ActiveRecord で次/前のレコードを取得しようとしています。レコードは、「updated_at」列の順序に従って取得する必要があります。

モデルの名前は「Youtube」です。そして、次のコンソールのように、このコードは正しいレコードを取得できませんでした。updated_at が常に一意であるとは限らず、一部のレコードが同じタイムスタンプを持つ可能性があるため、コードの考え方が悪いように思われます。

次/前のレコードを正しい方法で取得するにはどうすればよいですか?

コンソールは以下のように述べています。

[57] pry(main)> Youtube.find(1000)
  Youtube Load (0.5ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 1000]]
=> #<Youtube id: 1000, author_id: 2, category_label: nil, generated_by: 1, title: "Is Kenya Mall Shooting Over? Were Americans Among A...", video_id: "4T1szQIQcNI", created_at: "2013-09-30 18:31:21", updated_at: "2013-10-27 02:19:56", subtitles: nil>
[58] pry(main)> Youtube.find(1000).next
  Youtube Load (0.6ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 1000]]
Sun, 27 Oct 2013 02:19:56 UTC +00:00
  Youtube Load (256.6ms)  SELECT "youtubes".* FROM "youtubes" WHERE (updated_at > '2013-10-27 02:19:56.593969') ORDER BY updated_at DESC LIMIT 1
=> #<Youtube id: 67003, author_id: 75, category_label: nil, generated_by: 1, title: "Jewelry Photography : Lenses for Jewelry Photograph...", video_id: "NqA7OZL4tzw", created_at: "2013-10-09 17:18:53", updated_at: "2013-10-28 02:17:33", subtitles: nil>
[59] pry(main)> Youtube.find(1000).previous
  Youtube Load (0.6ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 1000]]
Sun, 27 Oct 2013 02:19:56 UTC +00:00
  Youtube Load (56.3ms)  SELECT "youtubes".* FROM "youtubes" WHERE (updated_at < '2013-10-27 02:19:56.593969') ORDER BY updated_at DESC LIMIT 1
=> #<Youtube id: 999, author_id: 8, category_label: nil, generated_by: 1, title: "Authors@Google: Richard Moore, Ned Boulting, and Da...", video_id: "4SCzfuJAyJw", created_at: "2013-09-30 18:31:21", updated_at: "2013-10-27 02:19:55", subtitles: nil>

YouTube には、次の default_scope があります。状況によっては変更される可能性もありますが、このコードは現状維持のために残していただければと思います。

  default_scope order('updated_at DESC')

Youtube Model の試用コードは以下のとおりです。

  scope :next, lambda{|updated_at| where("updated_at > ?",
    updated_at).order("updated_at DESC")}
  scope :previous, lambda {|updated_at| where("updated_at < ?",
    updated_at).order("updated_at DESC")}

...

  def next
    self.class.next(updated_at).first
  end 

  def previous                 
    self.class.previous(updated_at).first
  end
4

3 に答える 3

6

私は試行錯誤しましたが、以下が解決策の1つであることがわかりました。

コードはこちらです。

  def next
    self.class.unscoped.where("updated_at <= ? AND id != ?", updated_at, id).order("updated_at DESC").first
  end

  def previous
    self.class.unscoped.where("updated_at >= ? AND id != ?", updated_at, id).order("updated_at ASC").first
  end

テストはここにあります。

[210] pry(main)> Youtube.find(100)
  Youtube Load (0.8ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 100]]
=> #<Youtube id: 100, author_id: 5, category_label: nil, generated_by: 1, title: "Woman's Profane Dunkin Donuts Rant Goes Viral", video_id: "-aqN7KdWgQE", created_at: "2013-09-30 18:19:42", updated_at: "2013-10-27 00:47:37", subtitles: nil>
[211] pry(main)> Youtube.find(100).next
  Youtube Load (0.7ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 100]]
  Youtube Load (95.9ms)  SELECT "youtubes".* FROM "youtubes" WHERE (updated_at <= '2013-10-27 00:47:37.241076' AND id != 100) ORDER BY updated_at DESC LIMIT 1
=> #<Youtube id: 99, author_id: 6, category_label: nil, generated_by: 1, title: "Editing physical locations in Google Maps", video_id: "-amPC4fcY0U", created_at: "2013-09-30 18:19:42", updated_at: "2013-10-27 00:47:36", subtitles: nil>
[212] pry(main)> Youtube.find(100).next.previous
  Youtube Load (0.7ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 100]]
  Youtube Load (68.8ms)  SELECT "youtubes".* FROM "youtubes" WHERE (updated_at <= '2013-10-27 00:47:37.241076' AND id != 100) ORDER BY updated_at DESC LIMIT 1
  Youtube Load (79.5ms)  SELECT "youtubes".* FROM "youtubes" WHERE (updated_at >= '2013-10-27 00:47:36.162671' AND id != 99) ORDER BY updated_at ASC LIMIT 1
=> #<Youtube id: 100, author_id: 5, category_label: nil, generated_by: 1, title: "Woman's Profane Dunkin Donuts Rant Goes Viral", video_id: "-aqN7KdWgQE", created_at: "2013-09-30 18:19:42", updated_at: "2013-10-27 00:47:37", subtitles: nil>
[213] pry(main)> Youtube.find(100) === Youtube.find(100).next.previous
  Youtube Load (0.8ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 100]]
  Youtube Load (4.8ms)  SELECT "youtubes".* FROM "youtubes" WHERE "youtubes"."id" = $1 ORDER BY updated_at DESC LIMIT 1  [["id", 100]]
  Youtube Load (99.7ms)  SELECT "youtubes".* FROM "youtubes" WHERE (updated_at <= '2013-10-27 00:47:37.241076' AND id != 100) ORDER BY updated_at DESC LIMIT 1
  Youtube Load (79.6ms)  SELECT "youtubes".* FROM "youtubes" WHERE (updated_at >= '2013-10-27 00:47:36.162671' AND id != 99) ORDER BY updated_at ASC LIMIT 1
=> true
于 2013-10-28T03:40:17.613 に答える
0

私はこのようにします（より良い方法かもしれません）...

scope :in_order, lambda{ order("updated_at DESC, id ASC") }

def next
  self.class.in_order.where("updated_at >= ? AND id != ?", updated_at, id).first
end 

def previous
  self.class.in_order.where("updated_at <= ? AND id != ?", updated_at, id).first
end
于 2013-10-28T02:40:49.860 に答える