返回符合特定条件的每个组的记录的10％

Question

我正在尝试查询具有多个组的表以获得满足特定条件的每个组中10％的系统。

这里的数据：

create table tablename as 
select '4FH6V1' as ComputerName, 'AZ1' as Location, '3093' as Rating from dual union all 
select '0GLYQ1' as ComputerName, 'AZ1' as Location, '3093' as Rating from dual union all 
select 'P191R1' as ComputerName, 'AZ1' as Location, '3093' as Rating from dual union all 
select '7CMJ02' as ComputerName, 'AZ3' as Location, '3392' as Rating from dual union all 
select '8W2QS1' as ComputerName, 'AZ4' as Location, '3093' as Rating from dual union all 
select 'K9CHX1' as ComputerName, 'AZ7' as Location, '3192' as Rating from dual union all 
select '3XZNS1' as ComputerName, 'AZ7' as Location, '3093' as Rating from dual union all 
select '79RGX1' as ComputerName, 'AZ9' as Location, '3192' as Rating from dual union all 
select '02BR22' as ComputerName, 'AZ3' as Location, '2593' as Rating from dual 
; 

| ComputerName | Location | Rating | 
|--------------|----------|--------| 
| 4FH6V1  | AZ1  | 3093 | 
| 0GLYQ1  | AZ1  | 3093 | 
| P191R1  | AZ1  | 3093 | 
| 7CMJ02  | AZ3  | 3392 | 
| 8W2QS1  | AZ4  | 3093 | 
| K9CHX1  | AZ7  | 3192 | 
| 3XZNS1  | AZ7  | 3093 | 
| 79RGX1  | AZ9  | 3192 | 
| 02BR22  | AZ3  | 2593 | 

还有更多的记录是这样的表格。例如，我需要找到计算机名称的10％，在有3093.

Answer 1

差饷每个位置如果您使用的是SQL服务器，你可以在你的select语句中使用PERCENT：

SELECT TOP 10 PERCENT * FROM tablename WHERE RATING=3093; 

Answer 2

如果您正在使用SQL Server，您可以使用apply：

select tt.* 
from (select distinct location from t where t.rating = 3093) l apply 
    (select top (10) PERCENT t.* 
     from t 
     where t.location = l.location and t.rating = 3093 
    ) tt 

Answer 3

这应该在Oracle工作，虽然样本数据是不是足够大，以获得10％的样本...

create table tablename as 
select '4FH6V1' as ComputerName, 'AZ1' as Location, '3093' as Rating from dual union all 
select '0GLYQ1' as ComputerName, 'AZ1' as Location, '3093' as Rating from dual union all 
select 'P191R1' as ComputerName, 'AZ1' as Location, '3093' as Rating from dual union all 
select '7CMJ02' as ComputerName, 'AZ3' as Location, '3392' as Rating from dual union all 
select '8W2QS1' as ComputerName, 'AZ4' as Location, '3093' as Rating from dual union all 
select 'K9CHX1' as ComputerName, 'AZ7' as Location, '3192' as Rating from dual union all 
select '3XZNS1' as ComputerName, 'AZ7' as Location, '3093' as Rating from dual union all 
select '79RGX1' as ComputerName, 'AZ9' as Location, '3192' as Rating from dual union all 
select '02BR22' as ComputerName, 'AZ3' as Location, '2593' as Rating from dual 
; 

with cte1 as (
select 
    x.ComputerName 
    ,x.Location 
    ,x.Rating 
    ,count(1) over (partition by Location) as location_total 
    ,row_number() over (partition by Location order by ComputerName) as location_position 
    ,row_number() over (partition by Location order by ComputerName)/count(1) over (partition by Location) as location_pct 
from tablename x 
where x.Rating = 3093 
) 
select * 
from cte1 
where location_pct <= 0.1