DB2——sum over partition by 的用法
参考的博文出处：http://www.cnblogs.com/luhe/p/4155612.html，对博文进行了修改新增，修改了错误的地方
之前用过row_number()，rank()等排序与over( partition by ... ORDER BY ...)，这两个比较好理解: 先分组，然后在组内排名。
今天突然碰到sum(...) over( partition by ... ORDER BY ... )，居然搞不清除怎么执行的，所以查了些资料，做了下实操。
1. 从最简单的开始
sum(...) over( )，对所有行求和
sum(...) over( order by ... )，和 = 第一行 到 与当前行同序号行的最后一行的所有值求和，文字不太好理解，请看下图的算法解析。
with aa as
(
SELECT 1 a,1 b, 3 c FROM dual union
SELECT 2 a,2 b, 3 c FROM dual union
SELECT 3 a,3 b, 3 c FROM dual union
SELECT 4 a,4 b, 3 c FROM dual union
SELECT 5 a,5 b, 3 c FROM dual union
SELECT 6 a,5 b, 3 c FROM dual union
SELECT 7 a,2 b, 3 c FROM dual union
SELECT 8 a,2 b, 8 c FROM dual union
SELECT 9 a,3 b, 3 c FROM dual
)
SELECT a,b,c,
sum(c) over(order by b) sum1,--有排序，求和当前行所在顺序号的C列所有值
sum(c) over() sum2--无排序，求和 C列所有值
from aa
补充 by 松门一枝花：
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WITH aa AS
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( SELECT 1 a,1 b, 3 c FROM dual
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UNION
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SELECT 2 a,2 b, 3 c FROM dual
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UNION
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SELECT 3 a,3 b, 3 c FROM dual
-
UNION
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SELECT 4 a,4 b, 3 c FROM dual
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UNION
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SELECT 5 a,5 b, 3 c FROM dual
-
UNION
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SELECT 6 a,5 b, 3 c FROM dual
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UNION
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SELECT 7 a,2 b, 3 c FROM dual
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UNION
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SELECT 8 a,2 b, 8 c FROM dual
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UNION
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SELECT 9 a,3 b, 3 c FROM dual
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)
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SELECT a,
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b,
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c,
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SUM(c) over(order by a) sum1,--有排序，求和当前行所在顺序号的C列所有值--【博主新增的】
-
SUM(c) over(order by b) sum2,--有排序，求和当前行所在顺序号的C列所有值
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SUM(c) over() sum3 FROM aa order by a; --无排序，求和 C列所有值
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2. 与 partition by 结合
sum(...) over( partition by... )，同组内所行求和
sum(...) over( partition by... order by ... )，同第1点中的排序求和原理，只是范围限制在组内
with aa as
(
SELECT 1 a,1 b, 3 c FROM dual union
SELECT 2 a,2 b, 3 c FROM dual union
SELECT 3 a,3 b, 3 c FROM dual union
SELECT 4 a,4 b, 3 c FROM dual union
SELECT 5 a,5 b, 3 c FROM dual union
SELECT 6 a,5 b, 3 c FROM dual union
SELECT 7 a,2 b, 3 c FROM dual union
SELECT 7 a,2 b, 8 c FROM dual union
SELECT 9 a,3 b, 3 c FROM dual
)
SELECT a,b,c,sum(c) over( partition by b ) partition_sum,
sum(c) over( partition by b order by a desc) partition_order_sum
FROM aa;
