SQL在表的末尾添加行和列，以顯示平均

Question

我提供了以下數據集：

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 

從數據集，我已經計算概率過去幾年的某些天。這是我在SQL已經寫道：

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 

運行查詢後，結果如下：

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 

我在這裏的問題是，我不能夠增加一列和行顯示平均值。這是我試圖得到的最終結果：

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 

感謝您的時間和事先幫助。

Answer 1

只需對所有列進行求和並將其除以您想要求和的列數。像這樣的東西。

;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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