SQL在表尾添加行列显示平均值答案

【问题标题】：SQL add a row and column at the end of the table to display averageSQL在表尾添加行列显示平均值
【发布时间】：2015-03-14 02:49:51
【问题描述】：

我得到了以下数据集：

ID   STATION ID   DATE         ELEMENT ID   00     01     02     03......23
1    6106000      2009-05-11   110          2550   900    1050   1550    3000
2    6106000      2009-05-12   110          380    1400   1550   5090    200
3    6106000      2009-05-13   110          500    1250   800    3550    4020

从数据集中，我计算了过去几年中某些日子的概率。这是我用 SQL 写的：

SELECT * FROM
(SELECT 
DatePart(year, date) AS 'Date',
Format([DATE],'MMM dd') AS 'Day',
Probability = Cast(Round(((
IIf([00]>1410,1,IIf([00]=0,1,0)) + 
IIf([01]>1410,1,IIf([01]=0,1,0)) + 
IIf([02]>1410,1,IIf([02]=0,1,0)) + 
IIf([03]>1410,1,IIf([03]=0,1,0)) + 
IIf([04]>1410,1,IIf([04]=0,1,0)) + 
IIf([05]>1410,1,IIf([05]=0,1,0)) + 
IIf([06]>1410,1,IIf([06]=0,1,0)) + 
IIf([07]>1410,1,IIf([07]=0,1,0)) + 
IIf([08]>1410,1,IIf([08]=0,1,0)) + 
IIf([09]>1410,1,IIf([09]=0,1,0)) + 
IIf([10]>1410,1,IIf([10]=0,1,0)) + 
IIf([11]>1410,1,IIf([11]=0,1,0)) + 
IIf([12]>1410,1,IIf([12]=0,1,0)) + 
IIf([13]>1410,1,IIf([13]=0,1,0)) + 
IIf([14]>1410,1,IIf([14]=0,1,0)) + 
IIf([15]>1410,1,IIf([15]=0,1,0)) + 
IIf([16]>1410,1,IIf([16]=0,1,0)) + 
IIf([17]>1410,1,IIf([17]=0,1,0)) + 
IIf([18]>1410,1,IIf([18]=0,1,0)) + 
IIf([19]>1410,1,IIf([19]=0,1,0)) + 
IIf([20]>1410,1,IIf([20]=0,1,0)) + 
IIf([21]>1410,1,IIf([21]=0,1,0)) + 
IIf([22]>1410,1,IIf([22]=0,1,0)) + 
IIf([23]>1410,1,IIf([23]=0,1,0)))/24.0)*100.0,0) AS int)
FROM 
ON_2
WHERE 
((([ELEMENT ID])=110) AND 
(([STATION ID])='6106000') AND 
((Day([DATE])) BETWEEN 1 AND 15) AND 
((DatePart("m",[DATE]))=12))) AS BaseData
PIVOT 
(MAX([Probability]) 
FOR [DATE] 
IN ([2000],[2001],[2002],[2003],[2004],[2005],[2006],[2007],[2008],[2009],[2010],[2011])) AS PivotTable

运行查询后，结果如下：

Day     2000  2001  2002  2003....2011
Dec 01  100   19    20    68......95
Dec 02  56    33    80    77......35
Dec 03  90    52    39    10......45
.
.
Dec 15  58    43    70    25......99

我的问题是我无法添加一列和一行来显示平均值。这是我想要得到的最终结果：

Day     2000  2001  2002  2003....2011  Avg
Dec 01  100   19    20    68......95    60
Dec 02  56    33    80    77......35    88
Dec 03  90    52    39    10......45    48
.
.
Dec 15  58    43    70    25......99    55
Avg     75    40    55    33......58

提前感谢您的时间和帮助。

【问题讨论】：

标签： sql-server dataset pivot average

【解决方案1】：

只需对所有列求和，然后除以您要求和的列数。像这样的。

;with cte as
(
select * from result
)
SELECT isnull(Day,'Avg'),
       Avg([2000]) [2000],
       Avg([2001]) [2001],
       Avg([2002]) [2002],
       Avg([2003]) [2003],
       Avg([2011]) [2011],
       ( Avg([2000]) + Avg([2001]) + Avg([2002])
         + Avg([2003]) + Avg([2011]) ) / 5 as [AVG]
FROM   cte
GROUP  BY day WITH rollup

SQLFIDDLE DEMO

【讨论】：

感谢您的帮助，但我无法获得有意义的结果
@OscarDuran - 现在更新检查