给出每个员工每年薪水涨幅超过5000的员工编号emp_no

问题描述

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_data-from_data=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’);

来源：牛客网
链接：https://www.nowcoder.com/practice/eb9b13e5257744db8265aa73de04fd44?tpId=82&&tqId=29779&rp=1&ru=/ta/sql&qru=/ta/sql/question-ranking

Sql语句

思路：自连接

select a.emp_no, a.from_date, (a.salary - b.salary) as salary_growth
from salaries as a, salaries as b
where a.emp_no = b.emp_no
and strftime('%Y', a.to_date) - strftime('%Y', b.to_date) = 1
and salary_growth > 5000
order by salary_growth desc;