Rails Memoization 有很多关系

Question

是否可以通过对包含 has_many 关联的 SQL 查询的记忆来缓存实例变量？在下面的两个示例方法中， some_instance 方法将按预期进行缓存，但 some_other_instance 将缓存 Artist 而不是歌曲关联，这会导致每次运行该方法时都会进行 SQL 查询。

当方法返回“ActiveRecord::Associations::CollectionProxy”或“ActiveRecord::Relation”时，它似乎不会完全缓存。

class Artist < ApplicationRecord
    has_many :artist_songs
    has_many :songs, through: :artist_songs

    def self.some_instance
        @some_instance ||= find(10)
    end

    def self.some_other_instance
        @some_other_instance ||= find(10).songs
    end
end

Answer 1

some_instance 方法缓存find(10) 方法执行数据库查询并返回记录的结果。

但是some_other_instance 方法缓存的是一个关系对象，而不是数据库查询的结果。实际上，如果您在 Artist.some_other_instance; （note ; at the end）这样的 rails 控制台中运行，则不会有任何数据库查询，因为您不使用查询的结果。但是当你只运行Artist.some_other_instance 时就会出现查询，因为结果是用来打印的。

所以要真正缓存some_other_instance 方法，您需要添加to_a 之类的方法，例如：

def self.some_other_instance @some_other_instance ||= find(10).songs.to_a end

但在这种情况下，此方法不会是“可链接的”。

Answer 2

to_a 提供了缓存关系的解决方案。

在下面的日志中，SQL 查询需要 1883.4 毫秒，构建视图需要 623.4 毫秒。总时间为 15053 毫秒，这意味着在对象上添加 to_a 需要 12546.2 毫秒。如果我关闭 to_a，总时间会减少到大约 2000 毫秒。

构建对象需要这么长时间是正常的吗？

Started GET "/build" for 127.0.0.1 at 2018-10-14 16:35:30 -0700
Processing by ArtistsController#build as HTML
  Artist Load (208.4ms)  SELECT `artists`.`id`, `artists`.`name`, `artists`.`uri`, `artists`.`artwork_url`, updates.popularity, updates.created_at AS pop_created_at, reachupdates.reach, reachupdates.created_at AS reach_created_at FROM `artists` INNER JOIN `updates` ON `updates`.`artist_id` = `artists`.`id` INNER JOIN `reachupdates` ON `reachupdates`.`artist_id` = `artists`.`id` WHERE `artists`.`id` = 1
  Song Load (510.5ms)  SELECT songs.*, songupdates.* FROM `songs` INNER JOIN `artist_songs` ON `artist_songs`.`song_id` = `songs`.`id` INNER JOIN `artists` ON `artists`.`id` = `artist_songs`.`artist_id` INNER JOIN `songupdates` ON `songupdates`.`song_id` = `songs`.`id` WHERE `artists`.`id` = 1
  Playlist Load (1089.2ms)  SELECT playlists.*, playlistupdates.* FROM `playlists` INNER JOIN `playlist_songs` ON `playlist_songs`.`playlist_id` = `playlists`.`id` INNER JOIN `songs` ON `songs`.`id` = `playlist_songs`.`song_id` INNER JOIN `playlistupdates` ON `playlistupdates`.`playlist_id` = `playlists`.`id` LEFT OUTER JOIN `artist_songs` ON `artist_songs`.`song_id` = `songs`.`id` LEFT OUTER JOIN `artists` ON `artists`.`id` = `artist_songs`.`artist_id` WHERE `artists`.`id` = 1 AND `playlists`.`relevant` = 1 AND (playlistupdates.id IN (SELECT MAX(playlistupdates.id) AS id FROM playlistupdates GROUP BY playlistupdates.playlist_id)) ORDER BY `playlistupdates`.`followers` DESC
  Rendering artists/build.html.haml within layouts/application
  Rendered artists/build.html.haml within layouts/application (15.6ms)
  Rendered layouts/_head.html.haml (45.8ms)
  Rendered svg/_settings.html.erb (0.9ms)
  Rendered svg/_search.html.erb (0.4ms)
  Rendered layouts/_header_new.html.haml (73.3ms)
  Rendered shared/_autocomplete-box.html.haml (14.2ms)
Completed 200 OK in 15053ms (Views: 623.4ms | ActiveRecord: 1883.4ms)