回答編集履歴
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追記
answer
CHANGED
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@@ -44,4 +44,31 @@
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> 20.4 最大値
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> テーブルに含まれるフィールドの最大値を得るには、そのテーブルを持つクラスに対してmaximumメソッドを呼び出します。このメソッド呼び出しは以下のようなものになります。
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> Client.maximum("age")
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> ...
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> ...
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追記： 2016-02-04
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created_at で desc でsort して、limit をとったら、ruby 側で目的とする項目でソートするのは如何でしょう。
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```
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Frame number: 0/20
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[1] pry(main)> xx = Client.order("created_at desc").limit(6)
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Client Load (130.5ms) SELECT "clients".* FROM "clients" ORDER BY created_at desc LIMIT 6
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+----+------------+-----------+--------+--------------+-----------------+-----------------+
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| id | first_name | last_name | locked | orders_count | created_at | updated_at |
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+----+------------+-----------+--------+--------------+-----------------+-----------------+
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| 12 | Nick | | false | | 2015-12-27 0... | 2015-12-27 0... |
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| 11 | AndyY | | true | | 2015-12-27 0... | 2015-12-27 0... |
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| 10 | AndyX | | true | | 2015-12-27 0... | 2015-12-27 0... |
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| 9 | Andy | Smith | false | | 2015-12-27 0... | 2015-12-27 0... |
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| 8 | ussel | Smith | false | | 2015-12-27 0... | 2015-12-27 0... |
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| 7 | 安倍 | 次郎 | true | 3 | 2015-12-27 0... | 2015-12-27 0... |
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+----+------------+-----------+--------+--------------+-----------------+-----------------+
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6 rows in set
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[2] pry(main)> xx.to_a.sort{|a,b| a.orders_count.to_i <=> b.orders_count.to_i}.reverse[0]
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+----+------------+-----------+--------+--------------+-----------------+-----------------+
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| id | first_name | last_name | locked | orders_count | created_at | updated_at |
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+----+------------+-----------+--------+--------------+-----------------+-----------------+
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| 7 | 安倍 | 次郎 | true | 3 | 2015-12-27 0... | 2015-12-27 0... |
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+----+------------+-----------+--------+--------------+-----------------+-----------------+
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1 row in set
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[3] pry(main)>
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```
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