我正在使用数据集,并且我需要计算概率过去几年的某些天。以下是数据集的外观:
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
我无法在数据透视表中找到显示所需值的方法。这是我在SQL中的内容:
SELECT * FROM
(SELECT
Convert(char(15),[DATE],107) 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
正如您所看到的,我正在SELECT部分中执行操作集,其中包括IIf statements
和其他一些数学运算。获得我想要的值后,我将值分配给了'概率'这是我在 PIVOT部分中使用的值,但我的结果如下:
Day 2000 2001 2002 2003....2011
Dec 01 NULL NULL NULL NULL....NULL
Dec 02 NULL NULL NULL NULL....NULL
Dec 03 NULL NULL NULL NULL....NULL
.
.
Dec 15 NULL NULL NULL NULL....NULL
我已经测试了IIf计算和所有数学计算。我能够获得我想要的数字,但是当我将其翻译成数据透视表时,我看不到这些值,只能看到NULL。
非常感谢你的帮助,谢谢你的时间!
答案 0 :(得分:2)
你需要使用
datepart(year, date)
而不是
Convert(char(15),[DATE],107) AS 'Date'