mysql每日一题0718----- 每个公司的工资的中位数
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最近关注了一个公众号，该公众号主要围绕着数据库为技术主线展开，里面有每日一题
- T0716 表包含所有员工，其中有三列：员工 Id，公司名和薪水。
| ID | Company | Salary |
|---|---|---|
| 1 | A | 8341 |
| 2 | A | 9410 |
| 3 | A | 10050 |
| 4 | A | 15314 |
| 5 | A | 8451 |
| 6 | A | 9513 |
| 7 | B | 10005 |
| 8 | B | 13000 |
| 9 | B | 11540 |
| 10 | B | 10345 |
| 11 | B | 12210 |
| 12 | B | 9234 |
| 13 | C | 12000 |
| 14 | C | 8900 |
| 15 | C | 9000 |
| 16 | C | 10100 |
| 17 | C | 8000 |
- 现在 的需求是求每个公司的工资的中位数
需求结果如下：
| ID | Company | Salary |
|---|---|---|
| 6 | A | 9513 |
| 2 | A | 9410 |
| 9 | B | 11540 |
| 10 | B | 10345 |
| 15 | C | 9000 |
代码实现：
CREATE TABLE T0716
(
ID INT,
Company VARCHAR(10),
Salary INT
);
INSERT INTO T0716 VALUES
(1,'A',8341),
(2,'A',9410),
(3,'A',10050),
(4,'A',15314),
(5,'A',8451),
(6,'A',9513),
(7,'B',10005),
(8,'B',13000),
(9,'B',11540),
(10,'B',10345),
(11,'B',12210),
(12,'B',9234),
(13,'C',12000),
(14,'C',8900),
(15,'C',9000),
(16,'C',10100),
(17,'C',8000)
查询语句如下：
- 创建临时表
- 根据中位数的条件判断筛选出合适的记录
WITH temp AS (
SELECT a.*,ROW_NUMBER() over (PARTITION BY company ORDER BY Salary DESC) paiming,count(1) over (PARTITION BY company) num FROM t0716 a)
SELECT a.ID,a.Company,a.Salary FROM temp a WHERE paiming=round(num/2,0) OR (paiming=round(num/2)+1 AND num % 2=0)
