如何获得累计金额

时间：2010-01-22 21:12:14

标签： sql-server sql-server-2008 

declare  @t table
    (
        id int,
        SomeNumt int
    )

insert into @t
select 1,10
union
select 2,12
union
select 3,3
union
select 4,15
union
select 5,23


select * from @t

以上选择将返回以下内容。

id  SomeNumt
1   10
2   12
3   3
4   15
5   23

我如何获得以下

id  srome   CumSrome
1   10  10
2   12  22
3   3   25
4   15  40
5   23  63

16 个答案:

答案 0 :(得分：189)

select t1.id, t1.SomeNumt, SUM(t2.SomeNumt) as sum
from @t t1
inner join @t t2 on t1.id >= t2.id
group by t1.id, t1.SomeNumt
order by t1.id

SQL Fiddle example

<强>输出 

| ID | SOMENUMT | SUM |
-----------------------
|  1 |       10 |  10 |
|  2 |       12 |  22 |
|  3 |        3 |  25 |
|  4 |       15 |  40 |
|  5 |       23 |  63 |

编辑：这是一种适用于大多数数据库平台的通用解决方案。如果您的特定平台有更好的解决方案（例如，gareth's），请使用它！

答案 1 :(得分：161)

最新版本的SQL Server（2012）允许以下内容。

SELECT 
    RowID, 
    Col1,
    SUM(Col1) OVER(ORDER BY RowId ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Col2
FROM tablehh
ORDER BY RowId

或

SELECT 
    GroupID, 
    RowID, 
    Col1,
    SUM(Col1) OVER(PARTITION BY GroupID ORDER BY RowId ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Col2
FROM tablehh
ORDER BY RowId

这更快。分区版本在34秒内完成超过500万行。

感谢Peso，他对另一个答案中提到的SQL Team线程发表了评论。

答案 2 :(得分：21)

对于SQL Server 2012以上，它可能很简单：

TypeError: Cannot read property 'status' of undefined

因为SELECT id, SomeNumt, sum(SomeNumt) OVER (ORDER BY id) as CumSrome FROM @t 的{​​{1}}条款默认表示ORDER BY用于窗口框架（＆＃34;一般说明＆＃34;在https://msdn.microsoft.com/en-us/library/ms189461.aspx）

答案 3 :(得分：11)

CTE版本，只是为了好玩：

;
WITH  abcd
        AS ( SELECT id
                   ,SomeNumt
                   ,SomeNumt AS MySum
             FROM   @t
             WHERE  id = 1
             UNION ALL
             SELECT t.id
                   ,t.SomeNumt
                   ,t.SomeNumt + a.MySum AS MySum
             FROM   @t AS t
                    JOIN abcd AS a ON a.id = t.id - 1
           )
  SELECT  *  FROM    abcd
OPTION  ( MAXRECURSION 1000 ) -- limit recursion here, or 0 for no limit.

返回：

id          SomeNumt    MySum
----------- ----------- -----------
1           10          10
2           12          22
3           3           25
4           15          40
5           23          63

答案 4 :(得分：11)

让我们首先创建一个包含虚拟数据的表 - ＆gt; 

Create Table CUMULATIVESUM (id tinyint , SomeValue tinyint)

**Now let put some data in the table**

Insert Into CUMULATIVESUM

Select 1, 10 union 
Select 2, 2  union
Select 3, 6  union
Select 4, 10

在这里我加入了同桌（SELF Joining） 

Select c1.ID, c1.SomeValue, c2.SomeValue
From CumulativeSum c1,  CumulativeSum c2
Where c1.id >= c2.ID
Order By c1.id Asc

结果： 

ID  SomeValue   SomeValue
1   10          10
2   2           10
2   2            2
3   6           10
3   6            2
3   6            6
4   10          10
4   10           2
4   10           6
4   10          10

这里我们现在就把t2的Somevalue加起来，我们就得到了 

Select c1.ID, c1.SomeValue, Sum(c2.SomeValue) CumulativeSumValue
From CumulativeSum c1,  CumulativeSum c2
Where c1.id >= c2.ID
Group By c1.ID, c1.SomeValue
Order By c1.id Asc
  

FOR SQL SERVER 2012及更高版本（更好的表现）

Select c1.ID, c1.SomeValue, 
SUM (SomeValue) OVER (ORDER BY c1.ID )
From CumulativeSum c1
Order By c1.id Asc

期望的结果 

ID  SomeValue   CumlativeSumValue
1   10          10
2   2           12
3   6           18
4   10          28

Drop Table CumulativeSum

清除虚拟表格

答案 5 :(得分：5)

迟到的答案，但又显示了一种可能性......

使用CROSS APPLY逻辑可以更优化累积和生成。

效果优于INNER JOIN＆amp;分析实际查询计划时OVER Clause ... 

/* Create table & populate data */
IF OBJECT_ID('tempdb..#TMP') IS NOT NULL
DROP TABLE #TMP 

SELECT * INTO #TMP 
FROM (
SELECT 1 AS id
UNION 
SELECT 2 AS id
UNION 
SELECT 3 AS id
UNION 
SELECT 4 AS id
UNION 
SELECT 5 AS id
) Tab


/* Using CROSS APPLY 
Query cost relative to the batch 17%
*/    
SELECT   T1.id, 
         T2.CumSum 
FROM     #TMP T1 
         CROSS APPLY ( 
         SELECT   SUM(T2.id) AS CumSum 
         FROM     #TMP T2 
         WHERE    T1.id >= T2.id
         ) T2

/* Using INNER JOIN 
Query cost relative to the batch 46%
*/
SELECT   T1.id, 
         SUM(T2.id) CumSum
FROM     #TMP T1
         INNER JOIN #TMP T2
                 ON T1.id > = T2.id
GROUP BY T1.id

/* Using OVER clause
Query cost relative to the batch 37%
*/
SELECT   T1.id, 
         SUM(T1.id) OVER( PARTITION BY id)
FROM     #TMP T1

Output:-
  id       CumSum
-------   ------- 
   1         1
   2         3
   3         6
   4         10
   5         15

答案 6 :(得分：3)

Select *, (Select SUM(SOMENUMT) From @t S Where S.id <= M.id) From @t M

答案 7 :(得分：2)

这篇优秀文章提供了更快的CTE实施： http://weblogs.sqlteam.com/mladenp/archive/2009/07/28/SQL-Server-2005-Fast-Running-Totals.aspx

此线程中的问题可以表示如下：      

    DECLARE @RT INT
    SELECT @RT = 0

    ;
    WITH  abcd
            AS ( SELECT TOP 100 percent
                        id
                       ,SomeNumt
                       ,MySum
                       order by id
               )
      update abcd
      set @RT = MySum = @RT + SomeNumt
      output inserted.*

DECLARE @RT INT SELECT @RT = 0 ; WITH abcd AS ( SELECT TOP 100 percent id ,SomeNumt ,MySum order by id ) update abcd set @RT = MySum = @RT + SomeNumt output inserted.*

答案 8 :(得分：1)

创建表格后

select 
    A.id, A.SomeNumt, SUM(B.SomeNumt) as sum
    from @t A, @t B where A.id >= B.id
    group by A.id, A.SomeNumt

order by A.id

答案 9 :(得分：1)

上面（Pre-SQL12）我们看到这样的例子： - 

SELECT
    T1.id, SUM(T2.id) AS CumSum
FROM 
    #TMP T1
    JOIN #TMP T2 ON T2.id < = T1.id
GROUP BY
    T1.id

效率更高...... 

SELECT
    T1.id, SUM(T2.id) + T1.id AS CumSum
FROM 
    #TMP T1
    JOIN #TMP T2 ON T2.id < T1.id
GROUP BY
    T1.id

答案 10 :(得分：0)

试试这个

select 
    t.id,
    t.SomeNumt, 
    sum(t.SomeNumt) Over (Order by t.id asc Rows Between Unbounded Preceding and Current Row) as cum
from 
    @t t 
group by
    t.id,
    t.SomeNumt
order by
    t.id asc;

答案 11 :(得分：0)

试试这个：

CREATE TABLE #t(
 [name] varchar NULL,
 [val] [int] NULL,
 [ID] [int] NULL
) ON [PRIMARY]

insert into #t (id,name,val) values
 (1,'A',10), (2,'B',20), (3,'C',30)

select t1.id, t1.val, SUM(t2.val) as cumSum
 from #t t1 inner join #t t2 on t1.id >= t2.id
 group by t1.id, t1.val order by t1.id

答案 12 :(得分：0)

SQL解决方案，它结合了“无限制前进和当前行之间的行”＃34;和＆＃34; SUM＆＃34;完成了我想要实现的目标。 非常感谢你！

如果它可以帮助任何人，这是我的情况。每当制造商被发现为＆＃34; Some Maker＆＃34;时，我想在列中累积+1。 （例）。如果没有，则没有增量但显示先前的增量结果。

所以这段SQL：

SUM( CASE [rmaker] WHEN 'Some Maker' THEN  1 ELSE 0 END) 
OVER 
(PARTITION BY UserID ORDER BY UserID,[rrank] ROWS BETWEEN UNBOUNDED PRECEDING AND CURRENT ROW) AS Cumul_CNT

允许我得到这样的东西：

User 1  Rank1   MakerA      0  
User 1  Rank2   MakerB      0  
User 1  Rank3   Some Maker  1  
User 1  Rank4   Some Maker  2  
User 1  Rank5   MakerC      2
User 1  Rank6   Some Maker  3  
User 2  Rank1   MakerA      0  
User 2  Rank2   SomeMaker   1

上述说明：它开始计算＆＃34;某个制造商＆＃34; 0，找到了一些Maker，我们做+1。对于用户1，发现了MakerC，所以我们不做+1，而是将Some Maker的垂直计数固定为2直到下一行。 分区是由用户进行的，因此当我们更改用户时，累计计数会回到零。

我在工作，我不希望这个答案有任何优点，只要说谢谢并展示我的例子，以防有人处于同样的情况。我试图将SUM和PARTITION结合起来，但结合了惊人的语法＆＃34;在无限制的前进行和当前的行之间行驶＃34;完成任务。

谢谢！ Groaker

答案 13 :(得分：0)

使用跟随查询不使用任何类型的JOIN累积工资：

SELECT * , (
  SELECT SUM( salary ) 
  FROM  `abc` AS table1
  WHERE table1.ID <=  `abc`.ID
    AND table1.name =  `abc`.Name
) AS cum
FROM  `abc` 
ORDER BY Name

答案 14 :(得分：0)

您可以使用此简单查询进行渐进式计算：

select 
   id
  ,SomeNumt
  ,sum(SomeNumt) over(order by id ROWS between UNBOUNDED PRECEDING and CURRENT ROW) as CumSrome
from @t

答案 15 :(得分：0)

对于Ex：如果您有一个包含两列的表，其中第一列是ID，第二列是数字，并希望找出累积和。

SELECT ID,Number,SUM(Number)OVER(ORDER BY ID) FROM T
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