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SQL习题038
1
查找排除当前最大、最小salary之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
69462.5555555556
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' 作条件限制,导致挑选出来的是全表最大、最小salary,然后对除去这两个salary再作条件限制 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同。
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)
正确的逻辑应如下所示,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')
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