Question

我需要找到符合其销售目标的人员的员工ID，并根据其销售目标为每个员工获得多少奖金。

create table ta(
empid int(10),
sales int(10)
);

empid  - sales
101    - 9898
201    - 988
301    - 87988
401    - 88
501    - 202088

 create table tb(
 target int(10),
 bonus varchar(100));

Target  - Bonus
1000    - 25%
5000    - 50%
10000   - 70%
50000   - 85%
100000  - 100%

================================= 我已经尝试过

select ta.*
from   ta
    right outer join tb on ta.sales=tb.target
where sales <= bonus;

还有

select empid, sales, target, bonus
from ta 
    join tb on ta.empid=tb.bonus
where target < 100000

Answer 1

首先获取员工已达到的所有目标，然后获取最高目标：

select t.empid, t.target, tb.bonus from (
  select empid, max(target) target
  from (
    select *
    from ta inner join tb
    on tb.target <= ta.sales
  ) g 
  group by empid  
) t inner join tb on tb.target = t.target

请参见demo。
结果：

| empid | target | bonus |
| ----- | ------ | ----- |
| 101   | 5000   | 50%   |
| 301   | 50000  | 85%   |
| 501   | 100000 | 100%  |

Answer 2

您可以使用GROUP BY使用以下解决方案：

SELECT ta.empid, ta.sales, MAX(tb.target) AS target, CONCAT(MAX(REPLACE(tb.bonus, '%', '') + 0), '%') AS bonus
FROM ta LEFT JOIN tb ON tb.target <= ta.sales
GROUP BY ta.empid, ta.sales
ORDER BY ta.empid

demo on dbfiddle.uk

要获得所有员工，可以使用LEFT JOIN。如果您只想查看有奖金的员工，则必须使用INNER JOIN或HAVING上的条件，如下所示：

SELECT ta.empid, ta.sales, MAX(tb.target) AS target, CONCAT(MAX(REPLACE(tb.bonus, '%', '') + 0), '%') AS bonus
FROM ta LEFT JOIN tb ON tb.target <= ta.sales
GROUP BY ta.empid, ta.sales
HAVING MAX(tb.target) IS NOT NULL -- or just using INNER JOIN
ORDER BY ta.empid

如何根据员工的销售目标找到奖金？

2 个答案: