SQL練習 – group by, having, 集計関数 COUNT

Topic：

給与レコード数が 15 を超える従業員番号とそれに対応するレコード数を検索する

給与表 salaries の情報は以下のように:

+--------+--------+------------+------------+
| emp_no | salary | from_date  | to_date    |
+--------+--------+------------+------------+
|  10001 |  60117 | 1986-06-26 | 1987-06-26 |
|  10001 |  62102 | 1987-06-26 | 1988-06-25 |
|  10001 |  66074 | 1988-06-25 | 1989-06-25 |
|  10001 |  66596 | 1989-06-25 | 1990-06-25 |
|  10001 |  66961 | 1990-06-25 | 1991-06-25 |
|  10001 |  71046 | 1991-06-25 | 1992-06-24 |
|  10001 |  74333 | 1992-06-24 | 1993-06-24 |
|  10001 |  75286 | 1993-06-24 | 1994-06-24 |
|  10001 |  75994 | 1994-06-24 | 1995-06-24 |
|  10001 |  76884 | 1995-06-24 | 1996-06-23 |
|  10001 |  80013 | 1996-06-23 | 1997-06-23 |
|  10001 |  81025 | 1997-06-23 | 1998-06-23 |
|  10001 |  81097 | 1998-06-23 | 1999-06-23 |
|  10001 |  84917 | 1999-06-23 | 2000-06-22 |
|  10001 |  85112 | 2000-06-22 | 2001-06-22 |
|  10001 |  85097 | 2001-06-22 | 2002-06-22 |
|  10001 |  88958 | 2002-06-22 | 9999-01-01 |
|  10002 |  72527 | 1996-08-03 | 1997-08-03 |
+--------+--------+------------+------------+

問題：

給与レコード数が 15 を超える従業員番号とそれに対応するレコード数を検索する

上記の例の出力は次のようになります。

+--------+----+
| emp_no | t  |
+--------+----+
|  10001 | 17 |
+--------+----+

解答：

GROUP BY と集計関数 COUNT を使用して、給与の変化が 15 を超える従業員を取得します。

SELECT emp_no, COUNT(emp_no) AS t
FROM salaries
GROUP BY emp_no
HAVING t > 15;

DDLとDMLは以下に添付するので、好きな方はご利用ください。

drop table if exists  `salaries` ; 
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
INSERT INTO salaries VALUES(10001,60117,'1986-06-26','1987-06-26');
INSERT INTO salaries VALUES(10001,62102,'1987-06-26','1988-06-25');
INSERT INTO salaries VALUES(10001,66074,'1988-06-25','1989-06-25');
INSERT INTO salaries VALUES(10001,66596,'1989-06-25','1990-06-25');
INSERT INTO salaries VALUES(10001,66961,'1990-06-25','1991-06-25');
INSERT INTO salaries VALUES(10001,71046,'1991-06-25','1992-06-24');
INSERT INTO salaries VALUES(10001,74333,'1992-06-24','1993-06-24');
INSERT INTO salaries VALUES(10001,75286,'1993-06-24','1994-06-24');
INSERT INTO salaries VALUES(10001,75994,'1994-06-24','1995-06-24');
INSERT INTO salaries VALUES(10001,76884,'1995-06-24','1996-06-23');
INSERT INTO salaries VALUES(10001,80013,'1996-06-23','1997-06-23');
INSERT INTO salaries VALUES(10001,81025,'1997-06-23','1998-06-23');
INSERT INTO salaries VALUES(10001,81097,'1998-06-23','1999-06-23');
INSERT INTO salaries VALUES(10001,84917,'1999-06-23','2000-06-22');
INSERT INTO salaries VALUES(10001,85112,'2000-06-22','2001-06-22');
INSERT INTO salaries VALUES(10001,85097,'2001-06-22','2002-06-22');
INSERT INTO salaries VALUES(10001,88958,'2002-06-22','9999-01-01');
INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03');