Question

我想通过ActiveRecord获取下一个/上一个记录。记录应根据列的顺序检索。

模特的名字是＆＃39; Youtube＆＃39;。并且作为以下控制台，此代码无法获得正确的记录，我猜我的代码的想法似乎很糟糕，因为 updated_at并不总是唯一的所以一些记录可能具有相同的时间戳。

如何以正确的方式获得下一个/上一个记录？

控制台在下面说。

[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具有以下default_scope。虽然这可能会根据某些情况而改变，但我希望这些代码能够保持现有的行为。

default_scope order('updated_at DESC')

我的Youtube模型的试用代码如下。

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

Answer 1

我确实尝试过错误，发现以下是其中一个解决方案。

代码在这里。

  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

测试在这里。

[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

Answer 2

我写了一个自动生成此类查询的gem，order_query：

class YouTube < ActiveRecord::Base
  include OrderQuery
  order_query :order_display, [
    [:updated_at, :desc],
    [:id, :desc]
  ]
end

video = YouTube.find(42)
pos = video.order_display
video.next
video.after
video.position
video.prev
video.before

Answer 3

我会这样做（可能是更好的方式）......

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

在Ruby on Rails上获取ActiveRecord中的Next / Previous记录

3 个答案: