Question

参考我之前的问题：

Get count of items and their values in one column

Get count percent of a record in a single query

我如何获得间隔的百分比？这样的例子：

  ItemId        count          Percent 
 ------------------------------------    
   1-2            2              33.3    
   3-4            4              66.6

谢谢

Answer 1

您的Intervals表可能是SQL Server 2008中的TVP。

SELECT Intervals.ItemId,
       [count] = COUNT(MyTbl.ItemID),
       [Percent] = 100.0 * COUNT(MyTbl.ItemID) / SUM(COUNT(MyTbl.ItemID)) OVER()
FROM   (VALUES(NULL,0, 'Less than 1'),
              (1,2,'1-2'),
              (3,4,'3-4'),
              (6,NULL,'More than 4')) Intervals (Low, High, ItemId)
       LEFT JOIN (VALUES(1),
                        (1),
                        (3),
                        (4),
                        (4),
                        (4)) MyTbl(ItemID)
         ON ( MyTbl.ItemID BETWEEN ISNULL(Intervals.Low, -2147483648) AND
                                        ISNULL(Intervals.High, 2147483647) )
GROUP  BY Intervals.ItemId,
          Intervals.Low
ORDER  BY Intervals.Low

Answer 2

试试这个

select itemId,count(*),(count(*)/xx.Tot)*100 as Percent
from tableName a
join (select sum(count) as Tot from tableName) xx on 1=1
group by ItemID

使用单个查询获取间隔的百分比

2 个答案: