Question

我试图找出工人在工作2年后可以预期的平均工资增长百分比。

Table workers
+-----------+------------+
| worker_id | hire_date  |
+-----------+------------+
|  10001    | 1986-06-26 |
|  10002    | 1996-08-03 |
+-----------+------------+

Table salaries
+-----------+--------+------------+------------+
| worker_id | salary | from_date  | to_date    |
+-----------+--------+------------+------------+
|  10001    |  60117 | 1986-06-26 | 1987-06-26 |
|  10001    |  62102 | 1987-06-26 | 1988-06-25 |
|  10001    |  66074 | 1988-06-25 | 1989-06-25 |
|  10001    |  66596 | 1989-06-25 | 1990-06-25 |
|  10001    |  66961 | 1990-06-25 | 1991-06-25 |
|  10002    |  65828 | 1996-08-03 | 1997-08-03 |
|  10002    |  65909 | 1997-08-03 | 1998-08-03 |
|  10002    |  67534 | 1998-08-03 | 1999-08-03 |
|  10002    |  69366 | 1999-08-03 | 2000-08-02 |
+-----------+--------+------------+------------+

目前，我有这个查询，但我没有得到任何结果。你能救我吗？

SELECT YEAR(hire_date), YEAR(from_date), MAX(salary)-MIN(salary)
FROM salaries JOIN workers using(worker_id)
WHERE YEAR(hire_date)-YEAR(from_date) = 2;

Answer 1

你想要雇用某人的工资和两年后的工资。

基本上，你想要为每个工人两次查找薪水，然后执行一些算术：

select avg(s2.salary - s.salary)
from workers w join
     salaries s
     on s.worker_id = w.worker_id and s.start_date = w.hire_date join
     salaries s2
     on s2.worker_id = w.worker_id and
        w.start_date + interval 2 year >= s2.start_date and
        w.start_date + interval 2 year < s2.to_date;

如果你想增加一个百分比，那就更像是：

select avg( (s2.salary / s.salary) - 1)

Mysql：在匹配特定日期差异时计算增加/减少百分比

1 个答案: