好吧,我不确定这个问题的最佳解决方法是什么。但是,让我举个例子。我试图在这里实现的是在尝试按降序排列P2值时得到等于进位的P2值.10到1。
我有一张很大的桌子:
Category_Id Brand_Id Carry P2_0 P2_1 P2_3 ... P2_10
9 54 59 12 3 17 .
7 6 102 4 0 3 .
9 71 54 20 1 0 .
9 75 98 34 4 0 .
7 10 140 59 5 4 .
这是我的代码的主要逻辑:
SELECT CategoryCode,Brand_Id, (CASE
WHEN P2_10 > = Carry Then 'Error'
WHEN P2_10 + P2_9 > = Carry Then '10'
WHEN P2_10 + P2_9 + P2_8 > = Carry Then '9'
WHEN P2_10 + P2_9 + P2_8 + P2_7 >= Carry Then '8'
WHEN P2_10 + P2_9 + P2_8 + P2_7 + P2_6 >= Carry Then '7'
WHEN P2_10 + P2_9 + P2_8 + P2_7 + P2_6 + P2_5 > = Carry Then '6'
WHEN P2_10 + P2_9 + P2_8 + P2_7 + P2_6 + P2_5 + P2_4 > = Carry Then '5'
WHEN P2_10 + P2_9 + P2_8 + P2_7 + P2_6 + P2_5 + P2_4 + P2_3 > = Carry Then '4'
WHEN P2_10 + P2_9 + P2_8 + P2_7 + P2_6 + P2_5 + P2_4 + P2_3 + P2_2 > = Carry Then '3'
WHEN P2_10 + P2_9 + P2_8 + P2_7 + P2_6 + P2_5 + P2_4 + P2_3 + P2_2 + P2_1 > = Carry Then '2'
WHEN P2_10 + P2_9 + P2_8 + P2_7 + P2_6 + P2_5 + P2_4 + P2_3 + P2_2 + P2_1 + P2_0 > = Carry Then '1'
ELSE NULL END) As Threshold from BQ_15
现在这里的问题是对于brand_id 6,如果进位是106那么
P2_10(50) + P2_9(50) + P2_8(3) + P2_7(3) = Carry (106) gives the right result
but if P2_10 + P2_9 + P2_8 + P2_7 > Carry it has to go back to previous result, if in the previous result the new P2 was '0' it has to back further.
so if P2_10(50) + P2_9(50) + P2_8(2) + P2_7(0) + P2_6(30) > Carry (106) then it should skip P2_7 (because it is zero) and go to P2_8 (desired result) but for my code it goes to P2_7.
我知道我没有包含任何跳过'0'的内容,这就是我的整个问题所在,我如何在SQL中迭代我的代码将适用于这两种情况并获得所需的结果。
提前致谢
答案 0 :(得分:2)
使用交叉申请进行行范围计算。我拿了4个p2_xx列,根据需要扩展它。
from (
-- sample data
values (9,54,106, 50,50,2,0,30)
) hugeTable (Category_Id, Brand_Id, Carry, P2_10, P2_9, P2_8, P2_7, P2_6)
cross apply (
select Threshold = min(p2n)
from (
select p2n,
s = sum(p2val) over(order by p2n desc)
from (
values
(10, P2_10), (9, P2_9), (8, P2_8), (7, P2_7), (6, P2_6)
) t(p2n, p2val)
where p2val>0
) t
where s <= Carry
) t
我使用sum()over(),如果您使用的是2008或更早的版本
from (
-- sample data
values (9,54,106, 50,50,2,0,30)
) hugeTable (Category_Id, Brand_Id, Carry, P2_10, P2_9, P2_8, P2_7, P2_6)
cross apply (
select Threshold = min(p2n)
from (
select p2n,
s = (select sum(p2val)
from (
values
(10, P2_10), (9, P2_9), (8, P2_8), (7, P2_7), (6, P2_6)
) t2(p2n, p2val)
where t2.p2n>=t.p2n )
from (
values
(10, P2_10), (9, P2_9), (8, P2_8), (7, P2_7), (6, P2_6)
) t(p2n, p2val)
where p2val>0
) t
where s <= Carry
) t
答案 1 :(得分:1)
我怀疑您使用cross appy
在上一个答案中遇到任何问题。如果你确实有并发症,或者如果你想坚持你的原始案例表达,那么找到一个有效的表达并不是很难 - 它只是有点乱。以下是其中一个案例:
WHEN P2_10 + P2_9 + P2_8 + P2_7 >= Carry
THEN coalesce(
nullif(sgn(P2_8) * 8, 0),
nullif(sgn(P2_9) * 9, 0),
nullif(sgn(P2_10) * 10, 0),
-1
)
实际上,当你真的需要嵌套case
表达式时,这会不必要地复杂化。也许这是你没有意识到的事情:
WHEN P2_10 + P2_9 + P2_8 + P2_7 >= Carry
THEN case when P2_8 > 0 then 8 when P2_9 > 0 then 9 when P2_10 > 0 then 10 else -1 end
你有没有序列以全零开始然后跳过进位阈值:(0, 0, 0, 200)
?那些永远不会超过持有量的总和呢?我不相信其他答案涵盖了这种情况。