使用Partition的SQL中最高薪资和第五最高薪资的差异

Question

我有这样的数据：

CREATE TABLE salaries AS
SELECT * FROM ( VALUES
  ('US'    , 'A', 3935),
  ('US'    , 'B', 7805),
  ('US'    , 'C', 2302),
  ('US'    , 'D', 6772),
  ('US'    , 'E', 3173),
  ('US'    , 'F', 7739),
  ('Japan' , 'G', 3881),
  ('Japan' , 'H', 1158),
  ('Japan' , 'I', 2591),
  ('Japan' , 'J', 3758),
  ('Japan' , 'K', 8710),
  ('Japan' , 'L', 3376),
  ('France', 'M', 5768),
  ('France', 'N', 9466),
  ('France', 'O', 1750),
  ('France', 'P', 1049),
  ('France', 'Q', 3479),
  ('France', 'R', 5305)
) AS t(country,employee,salary);

为了找出每个国家的最高工资和第五最高工资之间的差别，我正在尝试以下办法：

select max(salary) over (partition by country) - rank(5) over (partition by country) 
from salaries

但是它抛出了以下错误：

“组内为有序集聚合函数所必需”

有人可以建议不使用任何连接的任何方法吗？

Answer 1

select      country
           ,max(salary) - max(case dr when 5 then salary end) as salary_diff

from       (select      country     
                       ,salary
                       ,dense_rank() over (partition by country order by salary desc) as dr

            from        salaries
            ) s

group by    country            

Answer 2

1. 创建一个定义所需内容的窗户 (按国家划分，按薪资降序)
2. 通过该窗口计算max(salary)
3. 从顶部计算nth_value (因为它们是按薪资递减排序的)

特别要注意的是，WINDOWS有范围和行。它们详细说明了如何在该窗口内执行计算。在这里，我们必须解除对WINDOW的绑定，以使nth_value()工作。通常情况下，它会计算所看到的一切，所以只有当它看到这一行时，nth_value才会起作用--但是我们可以让它在前面看到。

密码，

SELECT *
  , max(salary) OVER w1 - nth_value(salary,5) OVER w1 AS max_minus_fifth_highest
FROM foo
WINDOW w1 AS (
  PARTITION BY (country)
  ORDER BY SALARY desc
  ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING
)
ORDER BY country;

Sql Demo

Answer 3

select country, max_sal - salary
from (
     select country, salary, 
            max(salary) over (partition by country) max_sal,
           case when (row_number() over (partition by country
                                         order by salary desc)) = 5 
                then 1 
           end fifth
from table
) t where fifth is not null;