之前用過row_number(),rank()等排序與over( partition by ... ORDER BY ...),這兩個比較好理解: 先分組,然後在組內排名。
今天突然碰到sum(...) over( partition by ... ORDER BY ... ),居然搞不清除怎麼執行的,所以查了些資料,做了下實操。
1. 從最簡單的開始
sum(...) over( ),對所有行求和
sum(...) over( order by ... ),和 = 第一行 到 與當前行同序號行的最後一行的所有值求和,文字不太好理解,請看下圖的算法解析。
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,求和當前行所在順序號的C列所有值 sum(c) over() sum2--無排序,求和 C列所有值
2. 與 partition by 結合
sum(...) over( partition by... ),同組內所行求和
sum(...) over( partition by... order by ... ),同第1點中的排序求和原理,只是範圍限制在組內
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;