sum() over (order by )

sum(x) over( partition by y ORDER BY z ) 分析

sum(x) over (partition by y order by z)

求安照y分区，然后按z排序，连续加当前顺序号前面的数值 (求每个分区中，按照z的顺序累计求和)

a b

1 2

3 4

5 6

sum(b) over (order by a)

a b  sum

1 2  1

3 4  1+3

5 6 1+3+5

 

sum(b) over (order by a desc )

a b  sum

1 2  1+3+5

3 4  1+3

5 6  5

如果是sum（b）over（partition by a） 求的是按a分区，所有的和 

a b

1 3

1 4

2 3

3 1

3 4

结果是

a b  sum(b) (partition by a)

1 3 7

1 4 7

2 3 3

3 1 5

3 4 5

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

sum() over()

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;

 

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