SQL练习27：给出每个员工每年薪水涨幅超过5000的员工编号emp_no

题目链接：牛客网

题目描述
给出每个员工每年薪水涨幅超过5000的员工编号emp_no、薪水变更开始日期from_date以及薪水涨幅值salary_growth，并按照salary_growth逆序排列。
提示：在sqlite中获取datetime时间对应的年份函数为strftime('%Y', to_date)
(数据保证每个员工的每条薪水记录to_date-from_date=1年，而且同一员工的下一条薪水记录from_data=上一条薪水记录的to_data)

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：插入
INSERT INTO salaries VALUES(10001,52117,'1986-06-26','1987-06-26');
INSERT INTO salaries VALUES(10001,62102,'1987-06-26','1988-06-25');
INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03');
INSERT INTO salaries VALUES(10002,72527,'1997-08-03','1998-08-03');
INSERT INTO salaries VALUES(10002,72527,'1998-08-03','1999-08-03');
INSERT INTO salaries VALUES(10003,43616,'1996-12-02','1997-12-02');
INSERT INTO salaries VALUES(10003,43466,'1997-12-02','1998-12-02');

解法
根据题目的要求，可以连接两个工资表，然后根据根据时间和薪资涨幅筛选数据，最后根据salary_growth列进行排序。

SELECT s1.emp_no, s2.from_date, (s2.salary - s1.salary) salary_growth
FROM salaries s1 JOIN salaries s2
ON s1.emp_no = s2.emp_no
WHERE YEAR(s2.to_date) - YEAR(s1.to_date) = 1 AND s2.salary - s1.salary > 5000
ORDER BY salary_growth DESC