Question

我的表格如下：

name     marks
raja     88
ravi     88
karthik  99
praveen  99
vijay    70

在该表中，我假设表的等级为：

name      marks        rank 
karthirk    99           1
praveen     99           2
raja        88           3
ravi        88           4
vijay       77           5

我通过使用dense rank()函数

来实现

select  name,marks,dense_rank() over(  order by name ) as ranks
from std_D_D order by marks desc

但我需要的是从上面的表中我需要获得第二低的排名。我的意思是我想要的输出是：

ravi 88   4

我试图获得排名第二的排名，但我无法获得所有排名。我试过这个问题：

select max(a.ranks) as b from (
select  name,marks,dense_rank() over(  order by name ) as ranks
from std_D_D ) as a where a.ranks not in
(

select max(a.ranks) as b from (
select  name,marks,dense_rank() over(  order by name ) as ranks
from std_D_D ) as a )

Answer 1

您可以撤消排名顺序并选择第二个：

SELECT name, marks
FROM   (SELECT name, makrs, DENSE_RANK() OVER (ORDER BY name DESC) AS ranks
        FROM   std_D_D) t
WHERE  ranks = 2

Answer 2

Aswer对我的上述问题：

SELECT * FROM

(SELECT NAME,MARKS,dense_RANK() OVER(ORDER BY RANK DESC,name) AS TEMP_ORDER FROM

(SELECT *,DENSE_RANK() OVER(ORDER BY MARKS DESC,NAME) AS RANK FROM std_D_D) A
) B WHERE TEMP_ORDER=2;

Answer 3

首先我要说的是，如果您使用DENSE_RANK()，您将获得具有相同分数的人的相同排名，这与您的假设排名不同。

修改

根据OP的评论，添加了额外的ORDER BY name DESC。

示例数据

DECLARE @std_D_D table ( name VARCHAR(10), marks int ) insert into @std_D_D values ('raja', 88), ('ravi', 88), ('karthik', 99), ('praveen', 99), ('vijay', 70);

<强>查询

;WITH CTE AS ( SELECT name,marks,dense_rank() over(ORDER BY marks ASC,name DESC) as ranks from @std_D_D ) SELECT name,marks,ranks FROM CTE WHERE ranks = 2

此查询的输出将为您提供

ravi 88 2

使用密集秩函数的第二个最大值

3 个答案: