參考的博文出處：http://www.cnblogs.com/luhe/p/4155612.html，對博文進行了修改新增，修改了錯誤的地方
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，然后在組內排名。
今天突然碰到sum(...) over( partition by?... ORDER BY?... )，居然搞不清除怎么執行的，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后一行的所有值求和，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
from aa
補充 by 松門一枝花：
WITH aa AS( SELECT 1 a,1 b, 3 c FROM dualUNIONSELECT 2 a,2 b, 3 c FROM dualUNIONSELECT 3 a,3 b, 3 c FROM dualUNIONSELECT 4 a,4 b, 3 c FROM dualUNIONSELECT 5 a,5 b, 3 c FROM dualUNIONSELECT 6 a,5 b, 3 c FROM dualUNIONSELECT 7 a,2 b, 3 c FROM dualUNIONSELECT 8 a,2 b, 8 c FROM dualUNIONSELECT 9 a,3 b, 3 c FROM dual)
SELECT a,b,c,SUM(c) over(order by a) sum1,--有排序，求和當前行所在順序號的C列所有值--【博主新增的】SUM(c) over(order by b) sum2,--有排序，求和當前行所在順序號的C列所有值SUM(c) over() sum3 FROM aa order by a; --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，只是范圍限制在組內
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_sumFROM aa;
根。A.32B.20C.16D.30 - 試題答案網問答...)
