MySQL 查找同一组或分区中行的差
备注:测试数据库版本为MySQL 8.0
文章目录
- 一.需求
- 二.解决方案
-
- 2.1 标量子查询方法
- 2.2 窗口函数
一.需求
返回每个员工的deptno、ename和sal以及与同一部门(即deptno值相同)的员工间的sal之差，
该差值在当前员工及同部门内紧随其后聘用的员工间计算而来。
对于每个部门中最新聘用的员工，这个差值为"N/A"。
二.解决方案
需要通过标量子查询查找到 下一个员工的工资，然后再来进行计算
MySQl 8.0开始引入了lag 和lead等窗口函数，可以更为简单的访问上一行和下一行
2.1 标量子查询方法
通过两层标量子查询，获取到下一个员工的工资，然后再进行计算
select deptno,ename,hiredate,sal,
coalesce(cast(sal - next_sal as char(10)) ,'N/A') as diff
from (
select e.deptno,
e.ename,
e.hiredate,
e.sal,
(select min(sal) from emp d
where d.deptno = e.deptno
and d.hiredate =
( select min(hiredate) from emp d
where e.deptno = d.deptno
and d.hiredate > e.hiredate)) as next_sal
from emp e
) x;
测试记录
mysql>
mysql> select deptno,ename,hiredate,sal,
-> coalesce(cast(sal - next_sal as char(10)) ,'N/A') as diff
-> from (
-> select e.deptno,
-> e.ename,
-> e.hiredate,
-> e.sal,
-> (select min(sal) from emp d
-> where d.deptno = e.deptno
-> and d.hiredate =
-> ( select min(hiredate) from emp d
-> where e.deptno = d.deptno
-> and d.hiredate > e.hiredate)) as next_sal
-> from emp e
-> ) x;
+--------+--------+------------+---------+----------+
| deptno | ename | hiredate | sal | diff |
+--------+--------+------------+---------+----------+
| 20 | SMITH | 1980-12-17 | 800.00 | -2175.00 |
| 30 | ALLEN | 1981-02-20 | 1600.00 | 350.00 |
| 30 | WARD | 1981-02-22 | 1250.00 | -1600.00 |
| 20 | JONES | 1981-04-02 | 2975.00 | -25.00 |
| 30 | MARTIN | 1981-09-28 | 1250.00 | 300.00 |
| 30 | BLAKE | 1981-05-01 | 2850.00 | 1350.00 |
| 10 | CLARK | 1981-06-09 | 2450.00 | -2550.00 |
| 20 | SCOTT | 1987-06-13 | 3000.00 | N/A |
| 10 | KING | 1981-11-17 | 5000.00 | 3700.00 |
| 30 | TURNER | 1981-09-08 | 1500.00 | 250.00 |
| 20 | ADAMS | 1987-06-13 | 1100.00 | N/A |
| 30 | JAMES | 1981-12-03 | 950.00 | N/A |
| 20 | FORD | 1981-12-03 | 3000.00 | 1900.00 |
| 10 | MILLER | 1982-01-23 | 1300.00 | N/A |
+--------+--------+------------+---------+----------+
14 rows in set (0.00 sec)
2.2 窗口函数
MySQL8.0之后有窗口函数，写起来就会方便很多
select deptno,ename,hiredate,sal,
case when next_sal = 'N/A' then 'N/A' else cast(sal - next_sal as char(10)) end as diff
from
(
SELECT a.empno,
a.ename,
a.deptno,
a.hiredate,
a.sal,
lead(sal, 1, 'N/A') over w as 'next_sal'
FROM emp a
window w as (PARTITION BY a.deptno ORDER BY hiredate ASC)
) x;
测试记录
mysql> select deptno,ename,hiredate,sal,
-> case when next_sal = 'N/A' then 'N/A' else cast(sal - next_sal as char(10)) end as diff
-> from
-> (
-> SELECT a.empno,
-> a.ename,
-> a.deptno,
-> a.hiredate,
-> a.sal,
-> lead(sal, 1, 'N/A') over w as 'next_sal'
-> FROM emp a
-> window w as (PARTITION BY a.deptno ORDER BY hiredate ASC)
-> ) x;
+--------+--------+------------+---------+-------+
| deptno | ename | hiredate | sal | diff |
+--------+--------+------------+---------+-------+
| 10 | CLARK | 1981-06-09 | 2450.00 | -2550 |
| 10 | KING | 1981-11-17 | 5000.00 | 3700 |
| 10 | MILLER | 1982-01-23 | 1300.00 | N/A |
| 20 | SMITH | 1980-12-17 | 800.00 | -2175 |
| 20 | JONES | 1981-04-02 | 2975.00 | -25 |
| 20 | FORD | 1981-12-03 | 3000.00 | 0 |
| 20 | SCOTT | 1987-06-13 | 3000.00 | 1900 |
| 20 | ADAMS | 1987-06-13 | 1100.00 | N/A |
| 30 | ALLEN | 1981-02-20 | 1600.00 | 350 |
| 30 | WARD | 1981-02-22 | 1250.00 | -1600 |
| 30 | BLAKE | 1981-05-01 | 2850.00 | 1350 |
| 30 | TURNER | 1981-09-08 | 1500.00 | 250 |
| 30 | MARTIN | 1981-09-28 | 1250.00 | 300 |
| 30 | JAMES | 1981-12-03 | 950.00 | N/A |
+--------+--------+------------+---------+-------+
14 rows in set (0.00 sec)