牛客SQL练习第54题

题目描述

查找排除最大、最小salary之后的当前(to_date = ‘9999-01-01’ )员工的平均工资avg_salary。

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`));

如：

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,'2001-08-02','9999-01-01');
INSERT INTO salaries VALUES(10003,43699,'2000-12-01','2001-12-01');
INSERT INTO salaries VALUES(10003,43311,'2001-12-01','9999-01-01');
INSERT INTO salaries VALUES(10004,70698,'2000-11-27','2001-11-27');
INSERT INTO salaries VALUES(10004,74057,'2001-11-27','9999-01-01');

输出格式:

avg_salary
73292

解答：

where 后不能跟聚合函数，可以采用子查询得到最小最大工资

select avg(salary)
from salaries
where to_date='9999-01-01'
and salary not in (select max(salary) from salaries where to_date='9999-01-01')
and salary not in (select min(salary) from salaries where to_date='9999-01-01')

不用not in也行·：

select avg(salary) as avg_salary
from salaries
where salary>(select min(salary) from salaries  where to_date='9999-01-01') 
and salary<(SELECT max(salary) from salaries  where to_date='9999-01-01')
and to_date='9999-01-01'