sql - Oracle SQL - 如何使用 RANK() 或 DENSE_RANK() 和 ROW_NUMBER() 分析函数获取不同的行？

Question

我希望获得每个部门的前 3 名不同工资。我可以使用 RANK() 做到这一点或 DENSE_RANK()或 ROW_NUMBER()但是我的 table 上有一些工资相同的记录。

下面提到的是我的查询及其结果。

20部门工资前三名应该是6000、3000、2975。
但是有 2 名员工的薪水为 3000，他们都是 2 级。所以它给了我这个部门的 4 条记录(1 条记录为 1 级，2 条记录为 2 级记录，1 条记录为 3 级记录)。

请建议/建议如何获得每个部门的不同前三名工资。

询问:

SELECT * FROM (
SELECT EMPNO, DEPTNO, SAL, 
DENSE_RANK() over (partition by deptno order by sal DESC) as RANK,
row_number() over (partition by deptno order by sal DESC) as ROWNO
from EMP)
WHERE RANK <= 3;

结果:

Empno Deptno    Salary Rank   Rowno
---------------------------------------- 
7839    10      5000    1      1
7782    10      2450    2      2
7934    10      1300    3      3
7935    20      6000    1      1
7788    20      3000    2      2
7902    20      3000    2      3
7566    20      2975    3      4
7698    30      2850    1      1
7499    30      1600    2      2
7844    30      1500    3      3

Answer 1

如果您在 row_number 中获得更具体的信息, 与 partitioning by dept,salary那么你可以结合row_number和 dense_rank在这个查询中:

with data_row as 
( 
select 7839 as empno, 10 as deptno, 5000 as salary from dual union all
select 7782 as empno, 10 as deptno, 2450 as salary from dual union all
select 7934 as empno, 10 as deptno, 1300 as salary from dual union all
select 1111 as empno, 10 as deptno, 1111 as salary from dual union all
select 7935 as empno, 20 as deptno, 6000 as salary from dual union all
select 7788 as empno, 20 as deptno, 3000 as salary from dual union all
select 7902 as empno, 20 as deptno, 3000 as salary from dual union all
select 7566 as empno, 20 as deptno, 2975 as salary from dual union all
select 2222 as empno, 20 as deptno, 2222 as salary from dual union all
select 7698 as empno, 30 as deptno, 2850 as salary from dual union all
select 7499 as empno, 30 as deptno, 1600 as salary from dual union all
select 7844 as empno, 30 as deptno, 1500 as salary from dual union all
select 3333 as empno, 30 as deptno, 1333 as salary from dual
)
select *
from
(
select 
       deptno,
       salary,
       dense_rank() over (partition by deptno order by salary desc) as drank,
       row_number() over (partition by deptno, salary order by salary desc) as rowno             

from data_row
)
where drank <=3 and
      rowno =1