DB2——sum over partition by 的用法

Posted 2021-01-02 Im X

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：

 

1. WITH aa AS
2. ( SELECT 1 a,1 b, 3 c FROM dual
3. UNION
4. SELECT 2 a,2 b, 3 c FROM dual
5. UNION
6. SELECT 3 a,3 b, 3 c FROM dual
7. UNION
8. SELECT 4 a,4 b, 3 c FROM dual
9. UNION
10. SELECT 5 a,5 b, 3 c FROM dual
11. UNION
12. SELECT 6 a,5 b, 3 c FROM dual
13. UNION
14. SELECT 7 a,2 b, 3 c FROM dual
15. UNION
16. SELECT 8 a,2 b, 8 c FROM dual
17. UNION
18. SELECT 9 a,3 b, 3 c FROM dual
19. )
20. SELECT a,
21. b,
22. c,
23. 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】
24. 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
25. SUM(c) over() sum3 FROM aa order by a; --无排序，求和 C列所有值
26.  

 

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;