@(一只心中无码的程序员)专栏程序员
- ~
查找排除当前最大、最小salary以后的员工的平均工资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
));
输出格式:
avg_salary
69462.5555555556code
本题逻辑有问题,在挑选当前最大、最小salary时没加 to_date = '9999-01-01' 做条件限制,致使挑选出来的是全表最大、最小salary,而后对除去这两个salary再做条件限制 to_date = '9999-01-01' ,求平均薪水,此时求出的平均薪水与题目逻辑要求的不一样。 SELECT AVG(salary) AS avg_salary FROM salaries WHERE to_date = '9999-01-01' AND salary NOT IN (SELECT MAX(salary) FROM salaries) AND salary NOT IN (SELECT MIN(salary) FROM salaries) 正确的逻辑应以下所示,但在本题OJ系统中通不过: SELECT AVG(salary) AS 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')