当前行的累计值+前一行的总和

Question

如何从表中转换表中的列:

ColumnA   ColumnB
2           a
3           b
4           c
5           d
1           a

对此:

ColumnA          ColumnB
3                 a
6(=3+3)           b   
10(=4+3+3)        c   
15(=5+4+3+3)      d 

我很有兴趣看到esp.你会选择什么方法.

Answer 1

像这样:

;WITH cte
AS
(
   SELECT ColumnB, SUM(ColumnA) asum 
   FROM @t 
   gROUP BY ColumnB

), cteRanked AS
(
   SELECT asum, ColumnB, ROW_NUMBER() OVER(ORDER BY ColumnB) rownum
   FROM cte
) 
SELECT (SELECT SUM(asum) FROM cteRanked c2 WHERE c2.rownum <= c1.rownum),
  ColumnB
FROM cteRanked c1;

这应该给你:

ColumnA    ColumnB
3             a
6             b
10            c
15            d

这是一个现场演示

Answer 2

我通常会避免尝试这样做,但以下内容符合您的要求:

declare @T table (ColumnA int,ColumnB char(1))
insert into @T(ColumnA,ColumnB) values
(2    ,       'a'),
(3   ,        'b'),
(4  ,         'c'),
(5 ,          'd'),
(1,           'a')

;With Bs as (
    select distinct ColumnB from @T
)
select
    SUM(t.ColumnA),b.ColumnB
from
    Bs b
        inner join
    @T t
        on
            b.ColumnB >= t.ColumnB
group by
    b.ColumnB

结果:

            ColumnB
----------- -------
3           a
6           b
10          c
15          d

对于小数据集,这没关系.但是对于较大的数据集,请注意表的最后一行依赖于获取SUM原始表的整个内容.

Answer 3

尝试下面的脚本，

DECLARE @T TABLE(ColumnA INT, ColumnB VARCHAR(50));

INSERT INTO @T VALUES
    (2, 'a'),
    (3, 'b'),
    (4, 'c'),
    (5, 'd'),
    (1, 'a');

SELECT  SUM(ColumnA) OVER(ORDER BY ColumnB) AS ColumnA,ColumnB
FROM    (   SELECT  SUM(ColumnA) AS ColumnA,ColumnB
            FROM    @T  GROUP BY ColumnB )T

Answer 4

不确定这是否是最佳的,但是如何(SQL Fiddle):

SELECT x.A + COALESCE(SUM(y.A),0) ColumnA, x.ColumnB
FROM
(
    SELECT SUM(ColumnA) A, ColumnB
    FROM myTable
    GROUP BY ColumnB
) x
LEFT OUTER JOIN
(
    SELECT SUM(ColumnA) A, ColumnB
    FROM myTable
    GROUP BY ColumnB
) y ON y.ColumnB < x.ColumnB
GROUP BY x.ColumnB, x.A