How to get cumulative sum

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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

the above select returns me the following.

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

How do I get the following:

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

Output

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

Edit: this is a generalized solution that will work across most db platforms. When there is a better solution available for your specific platform (e.g., gareth's), use it!

A CTE version, just for fun:

;
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.

Returns:

id          SomeNumt    MySum
----------- ----------- -----------
1           10          10
2           12          22
3           3           25
4           15          40
5           23          63
Select 
    *, 
    (Select Sum(SOMENUMT) 
     From @t S 
     Where S.id <= M.id)
From @t M

You can use this simple query for progressive calculation :

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

For Ex: IF you have a table with two columns one is ID and second is number and wants to find out the cumulative sum.

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