SQL在table末尾加行加列显示平均值

Question

我获得了以下数据集：

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

根据数据集，我计算了过去几年中某些日子的概率。这是我在 SQL:

中写的

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

经过运行查询，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。这就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

感谢您的时间和事先的帮助。

Answer 1

只需对所有列求和，然后除以要求和的列数。像这样。

;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

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