MySQL - サンプル コードの詳細を GROUP BY グループ化して最大フィールド値を取得します:
ユーザーのログイン レコード情報をクエリする必要があるビジネス シナリオがあるとします。テーブル構造は次のとおりです。データ:
CREATE TABLE `tb` ( `id` int(11) NOT NULL AUTO_INCREMENT, `uid` int(11) NOT NULL, `ip` varchar(16) NOT NULL, `login_time` datetime, PRIMARY KEY (`id`), KEY (`uid`) );
table データ状況:
INSERT INTO tb SELECT null, 1001, '192.168.1.1', '2017-01-21 16:30:47'; INSERT INTO tb SELECT null, 1003, '192.168.1.153', '2017-01-21 19:30:51'; INSERT INTO tb SELECT null, 1001, '192.168.1.61', '2017-01-21 16:50:41'; INSERT INTO tb SELECT null, 1002, '192.168.1.31', '2017-01-21 18:30:21'; INSERT INTO tb SELECT null, 1002, '192.168.1.66', '2017-01-21 19:12:32'; INSERT INTO tb SELECT null, 1001, '192.168.1.81', '2017-01-21 19:53:09'; INSERT INTO tb SELECT null, 1001, '192.168.1.231', '2017-01-21 19:55:34';
+----+------+---------------+---------------------+ | id | uid | ip | login_time | +----+------+---------------+---------------------+ | 1 | 1001 | 192.168.1.1 | 2017-01-21 16:30:47 | | 2 | 1003 | 192.168.1.153 | 2017-01-21 19:30:51 | | 3 | 1001 | 192.168.1.61 | 2017-01-21 16:50:41 | | 4 | 1002 | 192.168.1.31 | 2017-01-21 18:30:21 | | 5 | 1002 | 192.168.1.66 | 2017-01-21 19:12:32 | | 6 | 1001 | 192.168.1.81 | 2017-01-21 19:53:09 | | 7 | 1001 | 192.168.1.231 | 2017-01-21 19:55:34 | +----+------+---------------+---------------------+
SELECT uid, max(login_time) FROM tb GROUP BY uid;
+------+---------------------+ | uid | max(login_time) | +------+---------------------+ | 1001 | 2017-01-21 19:55:34 | | 1002 | 2017-01-21 19:12:32 | | 1003 | 2017-01-21 19:30:51 | +------+---------------------+
-- 错误写法 SELECT uid, ip, max(login_time) FROM tb GROUP BY uid; -- 错误写法
サブクエリを書く:
<br/>
SELECT a.uid, a.ip, a.login_time FROM tb a WHERE a.login_time in ( SELECT max(login_time) FROM tb GROUP BY uid);
5.5.50:
SELECT a.uid, a.ip, a.login_time FROM tb a WHERE a.login_time = ( SELECT max(login_time) FROM tb WHERE a.uid = uid);
+----+--------------------+-------+------+---------------+------+---------+------+------+-------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+--------------------+-------+------+---------------+------+---------+------+------+-------------+ | 1 | PRIMARY | a | ALL | NULL | NULL | NULL | NULL | 7 | Using where | | 2 | DEPENDENT SUBQUERY | tb | ALL | uid | NULL | NULL | NULL | 7 | Using where | +----+--------------------+-------+------+---------------+------+---------+------+------+-------------+
3を直接書くと改善されます
joinに直接変更すると改善されますパフォーマンスが良い:
+----+--------------------+-------+------+---------------+------+---------+------------+------+-------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+--------------------+-------+------+---------------+------+---------+------------+------+-------------+ | 1 | PRIMARY | a | ALL | NULL | NULL | NULL | NULL | 7 | Using where | | 2 | DEPENDENT SUBQUERY | tb | ref | uid | uid | 4 | test.a.uid | 1 | NULL | +----+--------------------+-------+------+---------------+------+---------+------------+------+-------------+
SELECT a.uid, a.ip, a.login_time FROM (SELECT uid, max(login_time) login_time FROM tb GROUP BY uid ) b JOIN tb a ON a.uid = b.uid AND a.login_time = b.login_time;
注: 最小値をグループ化したい場合は、対応する関数とシンボルを変更するだけです。
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