SQL - 找到行

Question

之间的值我有2个表：

• Employee表2列Name和Sales
• Rewards 2列Bonus和Range

样品数据表：

Employee      Rewards 

| Name | Sales |    | Bonus | Range | 
+------+-------+    +-------+-------+ 
| John | 112 |    | 2  | 200 | 
| Mary | 201 |    | 3  | 300 | 
| Joe | 400 |    | 5  | 500 | 
| Jack | 300 | 

如果他的销售< = Rewards.Range，每位员工都应该从Rewords表中获得奖金。我想选择Employee.Name和Rewards.Bonus。

在这种情况下，结果应该是：

| Name | Bonus | 
+------+-------+ 
| John | 2  | 
| Mary | 3  | 
| Joe | 5  | 
| Jack | 3  | 

任何想法，这个SQL查询将是什么？

感谢， ZB

Answer 1

我建议这种方法。可能有一个或两个语法错误。

select name 
, (select bonus from rewards 
where range = 
(select min(range) 
from rewards 
where range >= sales) 
) 
from employee 

Answer 2

我会使用lag让每个范围的底部部分，并加入其雇员的表：

SELECT name, bonus 
FROM employee e 
JOIN (SELECT bonus, 
       range AS top_range, 
       COALESCE(LAG(range) OVER (ORDER BY bonus ASC), 0) AS bottom_range 
     FROM rewards) r ON e.sales BETWEEN r.bottom_range AND r.top_range 

Answer 3

;with OrderedBonuses as (
    select e.name, r.bonus, row_number() over (partition by e.name order by r.Range desc) as ord 
    from Employee e 
      JOIN Rewards r on e.Sales <= r.Range 
) 
select name, bonus 
from OrderedBonuses 
where ord = 1; 

Answer 4

我只是测试这一个，它取回权结果与您的测试表：

SELECT a.name, 
     MIN(b.bonus) AS bonus 
FROM db.employee a 
INNER JOIN db.rewards b 
ON a.sales <= b.range 
GROUP BY a.name;