Otemachi.rb #7で発表した資料です。 ActiveRecordにおけるサブクエリの書き方について紹介しています。

1. いまさらサブクエリ
2018.06.12 Otemachi.rb #7
Yuya Taki

2. self.inspect
[1] https://github.com/muramurasan/okuribito [2] http://poject.herokuapp.com/
➢ gemのロゴを書いたり…
➢ たまにQiitaの記事を書いたり…
➢ 新卒で某SIerにてスマートメータのヘッドエ
ンドシステムの開発に携わったあと、渋谷の
ベンチャー企業にてRuby on Railsを学び、
2016年10月エネチェンジに入社。
Name : Yuya Taki
GitHub : yuyasat
Qiita : yuyasat
[1]
若輩者ですので、何卒優しくご教授いた
だければと思います。
（commitはしていない）
➢ ○よ○よ風ゲームをReact.jsで実装し
たり
[2]

3. 今回のモデル
0
*
0
*
id | name
----+------------
1 | 多部未華子
2 | 佐津川愛美
3 | 新垣結衣
4 | 堀北真希
5 | 吉高由里子
6 | 悠城早矢
id | actress_id | year | title
----+------------+------+---------------------
1 | 2 | 2005 | 蝉しぐれ
2 | 1 | 2006 | 夜のピクニック
3 | 4 | 2005 | ALWAYS 三丁目の夕日
4 | 2 | 2012 | 忍道-SHINOBIDO-
5 | 2 | 2016 | 貞子vs伽椰子
6 | 4 | 2013 | 県庁おもてなし課
7 | 5 | 2013 | 真夏の方程式
id | movie_id | key
----+----------+------------
1 | 1 | 時代劇
2 | 1 | 子役
3 | 3 | 昭和
4 | 5 | ホラー
5 | 7 | ミステリー
6 | 7 | 夏
7 | 6 | 公務員
8 | 6 | 地方活性
9 | 1 | 夏
class Actress < ApplicationRecord
has_many :movies
has_many :tags, through: :movies
end
class Movie < ApplicationRecord
has_many :tags
belongs_to :actress
end
class Tag < ApplicationRecord
belongs_to :movie
end
actresses
movies
tags

4. サブクエリとは
➢ クエリ内のクエリ
SELECT
"movies".*
FROM
"movies"
WHERE
"movies"."actress_id" IN (
SELECT
"actresses"."id"
FROM
"actresses"
WHERE
"actresses"."name" = '堀北真希'
)
;
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id | actress_id | year | title
----+------------+------+---------------------
3 | 4 | 2005 | ALWAYS 三丁目の夕日
6 | 4 | 2013 | 県庁おもてなし課

5. moviesに紐づけられているactressを取得したい
➢ moviesテーブルと内部結合すればいい
Actress.joins(:movies)
SELECT
"actresses".*
FROM
"actresses"
INNER JOIN
"movies"
ON
"movies"."actress_id" = "actresses"."id"
;
id | name
----+------------
2 | 佐津川愛美
1 | 多部未華子
4 | 堀北真希
2 | 佐津川愛美
2 | 佐津川愛美
4 | 堀北真希
5 | 吉高由里子
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actresses
movies

6. 重複を排除するためにdistinctする
SELECT
DISTINCT "actresses".*
FROM
"actresses"
INNER JOIN
"movies"
ON
"movies"."actress_id" = "actresses"."id"
;
id | name
----+------------
2 | 佐津川愛美
1 | 多部未華子
4 | 堀北真希
5 | 吉高由里子
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*
0
*
Actress.joins(:movies).distinct

7. よく使いそうなのでscopeにする
class Actress < ApplictionRecord
scope :having_movies, -> {
joins(:movies).distinct
}
end
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8. scope内でjoinはあまり使いたくない
Tag.joins(movie: :actress).merge(Actress.having_movies)
SELECT
DISTINCT "tags".*
FROM
"tags"
INNER JOIN
"movies"
ON
"movies"."id" = "tags"."movie_id"
INNER JOIN
"actresses"
ON
"actresses"."id" = "movies"."actress_id"
INNER JOIN
"movies"
ON
"movies"."actress_id" = "actresses"."id"
;
ERROR: table name "movies" specified
more than once
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movieに紐づいているactressに紐づいているmovieのtagを取得したい（かなり作為的。例がよくない）

9. 回避策（サブクエリ）
SELECT
"tags".*
FROM
"tags"
INNER JOIN
"movies"
ON
"movies"."id" = "tags"."movie_id"
INNER JOIN
"actresses"
ON
"actresses"."id" = "movies"."actress_id"
WHERE
"actresses"."id" IN (
SELECT
DISTINCT "actresses"."id"
FROM
"actresses"
INNER JOIN
"movies"
ON
"movies"."actress_id" = "actresses"."id"
)
;
Tag.joins(movie: :actress).where(actresses: { id: Actress.having_movies })
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*
➢ 突然のテーブル名
○ モデルで記述したい
➢ サブクエリの中でもJOIN
➢ なんか匂う

10. やっぱりmergeを使いたい
Tag.joins(movie: :actress).merge(Actress.having_movies)
SELECT
"tags".*
FROM
"tags"
INNER JOIN
"movies"
ON
"movies"."id" = "tags"."movie_id"
INNER JOIN
"actresses"
ON
"actresses"."id" = "movies"."actress_id"
WHERE
"actresses"."id" IN (
SELECT
DISTINCT "movies"."actress_id"
FROM
"movies"
)
;
➢ 実行したいSQLはこれ
class Actress < ApplictionRecord
scope :having_movies, -> {
where(id: Movie.select("actress_id").distinct)
}
end
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11. 実行コスト（EXPLAIN）
class Actress < ApplictionRecord
scope :having_movies, -> {
joins(:movies).distinct
}
end
Tag.joins(movie: :actress).where(actresses: { id: Actress.having_movies })
class Actress < ApplictionRecord
scope :having_movies, -> {
where(id: Movie.select("actress_id").distinct)
}
end
Tag.joins(movie: :actress).merge(Actress.having_movies)
class Actress < ApplictionRecord
scope :having_movies, -> {
where(id: Movie.select("actress_id"))
}
end
Tag.joins(movie: :actress).merge(Actress.having_movies)
171.22..215.55
90.27..115.51
93.93..122.11
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12. （補足）もう少し現実的な例
SELECT
DISTINCT "tags".*
FROM
"tags"
INNER JOIN "movies" ON "movies"."id" = "tags"."movie_id"
INNER JOIN "actresses" ON "actresses"."id" = "movies"."actress_id"
INNER JOIN "actress_song_relationships" ON "actress_song_relationships"."actress_id" = "actresses"."id"
INNER JOIN
"songs"
ON
"songs"."id" = "actress_song_relationships"."song_id"
INNER JOIN
"songs"
ON
"songs"."id" = "movies"."theme_song_id"
WHERE
"songs"."name" = '世界が終わる夜に '
ORDER BY
"tags"."id"
;
「世界が終わる夜に」がすきな曲としてもつ女優が主演かつ、主題歌が「かざぐる
ま」の映画のタグを取得したい。
ERROR: table name "songs" specified more than once
Tag.joins(movie: :actress).merge(
Actress.joins(:favorite_songs).merge(Song.where(name: '世界が終わる夜に ')).distinct
).merge(
Movie.joins(:theme_song).merge(Song.where(name: 'かざぐるま ')).distinct
)
例が非現実的だっ
たので
ここをscopeにしたとする

13. （補足）回避策：サブクエリ
SELECT
DISTINCT "tags".*
FROM
"tags"
INNER JOIN "movies" ON "movies"."id" = "tags"."movie_id"
INNER JOIN "actresses" ON "actresses"."id" = "movies"."actress_id"
INNER JOIN "Actress_song_relationships" ON "actress_song_relationships"."actress_id" = "actresses"."id"
INNER JOIN "songs" ON "songs"."id" = "actress_song_relationships"."song_id"
WHERE
"songs"."name" = '世界が終わる夜に '
AND
"movies"."theme_song_id" IN (
SELECT
"songs"."id"
FROM
"songs"
WHERE
"songs"."name" = 'かざぐるま '
)
Tag.joins(movie: :actress).merge(
Actress.joins(:favorite_songs).merge(Song.where(name: '世界が終わる夜に ')).distinct
).merge(
Movie.where(theme_song: Song.where(name: 'かざぐるま '))
)
「世界が終わる夜に」がすきな曲としてもつ女優が主演かつ、主題歌が「かざぐる
ま」の映画のタグを取得したい。
片方をサブクエリにする

14. もう少し複雑なサブクエリの例
➢ 映画公開年の降順にActressを取り出したい
id | actress_id | year | title
----+------------+------+---------------------
1 | 2 | 2005 | 蝉しぐれ
2 | 1 | 2006 | 夜のピクニック
3 | 4 | 2005 | ALWAYS 三丁目の夕日
4 | 2 | 2012 | 忍道-SHINOBIDO-
5 | 2 | 2016 | 貞子vs伽椰子
6 | 4 | 2013 | 県庁おもてなし課
7 | 5 | 2013 | 真夏の方程式
id | name
----+------------
2 | 佐津川愛美
4 | 堀北真希
5 | 吉高由里子
1 | 多部未華子
0
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0
*
id | name
----+------------
1 | 多部未華子
2 | 佐津川愛美
3 | 新垣結衣
4 | 堀北真希
5 | 吉高由里子
6 | 悠城早矢
actresses
movies

15. SQLで記述
SELECT
actresses.*
FROM
actresses
INNER JOIN (
SELECT
"movies"."actress_id" AS actress_id,
MAX(year) AS max_year
FROM
"movies"
GROUP BY
"movies"."actress_id"
) AS movies_max_year
ON actresses.id = movies_max_year.actress_id
ORDER BY max_year DESC, actresses.id ASC
actress_id | max_year
------------+----------
5 | 2013
4 | 2013
2 | 2016
1 | 2006
id | name
----+------------
1 | 多部未華子
2 | 佐津川愛美
3 | 新垣結衣
4 | 堀北真希
5 | 吉高由里子
6 | 悠城早矢
id | name
----+------------
2 | 佐津川愛美
4 | 堀北真希
5 | 吉高由里子
1 | 多部未華子
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0
*
actressesmovies_max_year

16. ActiveRecordで記述（find_by_sql）
Actress.find_by_sql(%|
SELECT
actresses.*
FROM
actresses
INNER JOIN (
SELECT
"movies"."actress_id" AS actress_id,
MAX(year) AS max_year
FROM
"movies"
GROUP BY
"movies"."actress_id"
) AS movies_max_year
ON actresses.id = movies_max_year.actress_id
ORDER BY max_year DESC, actresses.id ASC
|)
➢ Arrayで出力
➢ Kaminariが使えない
○ Kaminari.paginate_arrayを使えばArrayでも使える
が、scopeで使いたい。
[#<Actress:0x007f8978525928
id: 2,
name: "佐津川愛美 ">,
#<Actress:0x007f8972e37968
id: 4,
name: "堀北真希">,
#<Actress:0x007f897851d0c0
id: 5,
name: "吉高由里子 ">,
#<Actress:0x007f897850fd58
id: 1,
name: "多部未華子 ">]
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17. ActiveRecordで記述（Arelを使う）
SELECT
actresses.*
FROM
actresses
INNER JOIN (
SELECT
"movies"."actress_id" AS actress_id,
MAX(year) AS max_year
FROM
"movies"
GROUP BY
"movies"."actress_id"
) AS movies_max_year
ON actresses.id = movies_max_year.actress_id
ORDER BY max_year DESC, actresses.id ASC
actress_at = Actress.arel_table
movie_at = Movie.arel_table
movies_max_year = movie_at.project(
Arel.sql(%|
"movies"."actress_id" as actress_id,
MAX(year) as max_year
|)
).group("movies.actress_id").as("movies_max_year")
join_conds = actress_at.join(
movies_max_year,
Arel::Nodes::InnerJoin
).on(
movies_max_year[:actress_id].eq(actress_at[:id])
).join_sources
Actress.joins(join_conds).order("movies_max_year.max_year DESC")
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18. ActiveRecordで記述（joinsの中にSQL直書き）
SELECT
actresses.*
FROM
actresses
INNER JOIN (
SELECT
"movies"."actress_id" AS actress_id,
MAX(year) AS max_year
FROM
"movies"
GROUP BY
"movies"."actress_id"
) AS movies_max_year
ON actresses.id = movies_max_year.actress_id
ORDER BY max_year DESC, actresses.id ASC
Actress.joins(%|
INNER JOIN (
SELECT
"movies"."actress_id" AS actress_id,
MAX(year) AS max_year
FROM
"movies"
GROUP BY
"movies"."actress_id"
) AS movies_max_year
ON actresses.id = movies_max_year.actress_id
|).order("max_year desc, actresses.id asc")
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19. scopeにして
class Actress < ApplicationRecord
scope :order_by_movie_year, -> {
actress_at = arel_table
movie_at = Movie.arel_table
movies_max_year = movie_at.project(
Arel.sql(%|
"movies"."actress_id" as actress_id,
MAX(year) as max_year
|)
).group("movies.actress_id").as("movies_max_year")
join_conds = actress_at.join(
movies_max_year,
Arel::Nodes::InnerJoin
).on(
movies_max_year[:actress_id].eq(actress_at[:id])
).join_sources
joins(join_conds).order("movies_max_year.max_year DESC")
}
end
pry> Actress.order_by_movie_year.class
=> Actress::ActiveRecord_Relation
scope :order_by_movie_year, -> {
joins(%|
INNER JOIN (
SELECT
"movies"."actress_id" AS actress_id,
MAX(year) AS max_year
FROM
"movies"
GROUP BY
"movies"."actress_id"
) AS movies_max_year
ON actresses.id = movies_max_year.actress_id
|).order("max_year DESC, actresses.id ASC")
}
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20. まとめ
➢ サブクエリを使ったほうが内部結合するよりもうまくscopeを定義できることが
ある
➢ where句でサブクエリを使うときは、
selectを使う
○ pluckを用いるとクエリが２つに
○ SQLのパフォーマンス次第では二つに分ける
のもアリ。
➢ 結合先にサブクエリを用いる場合
○ Arelを使う
○ joinsの中にSQLを直書きする
Actress.where(id: Movie.select("actress_id").distinct)
Actress.where(id: Movie.pluck(:actress_id).uniq)
「ActiveRecord サブクエリ」で検索するとArelの例が出てくるけど
joinsの中にSQLを書けば良いんじゃないかな。。

21. 資料
達人に学ぶDB設計 徹底指南書
Visual Representation of SQL Joins
ActiveRecordでサブクエリのJOIN
Arel でサブクエリ
ActiveRecordでサブクエリ（副問い合わせ）と内部結合

22. 余談

23. ご静聴ありがとうございました。