返回符合特定條件的每個組的記錄的10％

Question

我正在嘗試查詢具有多個組的表以獲得滿足特定條件的每個組中10％的系統。

這裏的數據：

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 | 

還有更多的記錄是這樣的表格。例如，我需要找到計算機名稱的10％，在有3093.

Answer 1

差餉每個位置如果您使用的是SQL服務器，你可以在你的select語句中使用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

這應該在Oracle工作，雖然樣本數據是不是足夠大，以獲得10％的樣本...

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