在下表中，我将计算列S：
Week L C1 C2 C3 C4 C5 C6 S
1 L1=3 258 141 140 21 51 197 S1=1000
2 7 291 261 64 275 72 249
3 1 287 164 99 222 149 172
4 3 163 193 A=200 95 170 51
5 4 193 11 271 B=300 199 26
6 L6=2 213 120 53 210 C=400 43 S6=507
7 5 36 208 243 163 126 D=100
8 3 174 56 282 274 214 222
9 1 63 295 M=222 153 136 173
10 4 191 114 195 N=8 28 188
11 3 125 293 38 56 O=277 11
作为通用公式，我想根据以下公式找到S列：
如果L> = 3，则S = C3在三周后+ C4在四周后+ C5在五周后+ C6在六周后。例如，S1 = A + B + c + D = 1000
如果L < 3，S = C3三周后+ C4四周后+ C5五周后。例如S6 = M + N + O = 507.
非常感谢你的帮助。
答案 0 :(得分：3)
第一个出现在我脑海中：
SELECT c1.*,
CASE WHEN c1.L >=3 THEN ISNULL(c2.C3,0) + ISNULL(c3.C4,0) + ISNULL(c4.C5,0) + ISNULL(c5.C6,0)
ELSE ISNULL(c2.C3,0) +ISNULL(c3.C4,0) + ISNULL(c4.C5,0) END as S
FROM YourTableName c1
LEFT JOIN YourTableName c2
ON c1.[Week] + 3 = c2.[week]
LEFT JOIN YourTableName c3
ON c2.[Week] + 1 = c3.[week]
LEFT JOIN YourTableName c4
ON c3.[Week] + 1 = c4.[week]
LEFT JOIN YourTableName c5
ON c4.[Week] + 1 = c5.[week]
输出：
Week L C1 C2 C3 C4 C5 C6 S
1 3 258 141 140 21 51 197 1000
2 7 291 261 64 275 72 249 829
3 1 287 164 99 222 149 172 430
4 3 163 193 200 95 170 51 841
5 4 193 11 271 300 199 26 474
6 2 213 120 53 210 400 43 507
7 5 36 208 243 163 126 100 251
8 3 174 56 282 274 214 222 38
9 1 63 295 222 153 136 173 0
10 4 191 114 195 8 28 188 0
11 3 125 293 38 56 277 11 0
修改
对于SQL Server 2012，您可以使用https://developer.paypal.com/docs/classic/adaptive-payments/integration-guide/APIntro/
SELECT *,
CASE WHEN L >= 3 THEN
LEAD(C3, 3, 0) OVER (ORDER BY [Week]) +
LEAD(C4, 4, 0) OVER (ORDER BY [Week]) +
LEAD(C5, 5, 0) OVER (ORDER BY [Week]) +
LEAD(C6, 6, 0) OVER (ORDER BY [Week])
ELSE
LEAD(C3, 3, 0) OVER (ORDER BY [Week]) +
LEAD(C4, 4, 0) OVER (ORDER BY [Week]) +
LEAD(C5, 5, 0) OVER (ORDER BY [Week])
END as S
FROM YourTableName
输出将是相同的。