[简单]oracle 按部门分组取每部门薪水最大的2条记录

       最近看到的一篇帖子，http://www.itpub.net/forum.php?mod=viewthread&tid=1865269，按部门分组取薪水最大的2条记录，原帖给出的答案：

      

select *
  from emp t
 where (select count(1) + 1
          from emp
         where nvl(deptno,0) = nvl(t.deptno,0)
           and nvl(sale,0) > nvl(t.sale,0)) <= 2

    结果为：

   

   

     Emp表按部门分组，按薪水排列结果如下：

     可以看到原帖的答案没有考虑到有null值情况，考虑空值的sql为：

    

select *
  from emp t
 where (select count(1) + 1
          from emp
         where nvl(deptno,0) = nvl(t.deptno,0)
           and nvl(sale,0) > nvl(t.sale,0)) <= 2
  order by deptno,sale desc nulls last

    结果为：

   

 

       或者使用分析函数：

      

select *
  from (select emp.*,
               dense_rank() over(partition by deptno order by sale desc nulls last) rn
          from emp)
 where rn <= 2;

   

     检验2个结果是否相同：

    

select t.empno
  from emp t
 where (select count(1) + 1
          from emp
         where nvl(deptno,0) = nvl(t.deptno,0)
           and nvl(sale, 0) > nvl(t.sale, 0)) <=2
minus
select t.empno
  from (select emp.*,
               dense_rank() over(partition by deptno order by sale desc nulls last) rn
          from emp) t
 where rn <= 2;

select t.empno
  from (select emp.*,
               dense_rank() over(partition by deptno order by sale desc nulls last) rn
          from emp) t
 where rn <= 2
minus
select t.empno
  from emp t
 where (select count(1) + 1
          from emp
         where nvl(deptno,0) = nvl(t.deptno,0)
           and nvl(sale, 0) > nvl(t.sale, 0)) <= 2

   

    欢迎提出更好的sql写法，谢谢。

 

     全文完。