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 FROM aa;--无排序,求和 C列所有值
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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