Select
Student_Name,
DENSE_RANK() OVER (ORDER BY dbms_random.VALUE(1,999999)) AS RBORV
From
Stu
从上面的sql我只需要提取前5%的排名(按顺序)..我应该在where子句中添加什么来这样做
有一种非常直接的方式:
Select TOP 5 PERCENT Student_Name,
DENSE_RANK() OVER (ORDER BY dbms_random.VALUE(1,999999)) AS RBORV
From Stu
Order by RBORV
您可以使用窗口函数执行此操作。使用您提供的代码:
select student_name
from (select Student_Name,
row_number() over (order by dbms_random.VALUE(1,999999)) AS RBORV,
count(*) over () as cnt
from Stu
) s
where row_number <= cnt * 0.05;
这将在Oracle中工作(看起来像您正在使用的代码)。在SQL Server中,您可以编写等效代码:
select student_name
from (select Student_Name,
row_number() over (order by newid()) AS RBORV,
count(*) over () as cnt
from Stu
) s
where row_number <= cnt * 0.05;
或者,更简单地说,如下:
select top 5 percent student_name
from stu
order by newid();