1    34
2    56
6    18
1    42
5    274
3    25
1    56
...

DECLARE @A table (Type int, Id int, Value int);
DECLARE @B table (Id int, Sum int);
DECLARE @Cursor cursor, @Id int, @Value int;

SET @Cursor = CURSOR FAST_FORWARD FOR SELECT Id, Value FROM @A;
OPEN @Cursor;

FETCH NEXT FROM @Cursor INTO @Id, @Value;
WHILE @@FETCH_STATUS = 0
    BEGIN
    IF NOT EXISTS (SELECT * FROM @B WHERE Id = @Id)
        INSERT @B (Id, Sum) VALUES (@Id, @Value);
    ELSE
        UPDATE @B SET Sum = Sum + @Value WHERE Id = @Id;

    FETCH NEXT FROM @Cursor INTO @Id, @Value;
    END

CLOSE @Cursor;
DEALLOCATE @Cursor;

create table A
(
  id int,
  value int
)

create table B
(
  id int,
  [sum] int
)

insert into A 
select 1,34 union all
select 2,56 union all
select 6,18 union all
select 1,42 union all
select 5,274 union all
select 3,25 union all
select 1,56 

--先删后插
delete from B where id in(select distinct id from A)

insert into B 
select id,SUM(value) from A group by id

--
select * from B

------解决方案--------------------

SQL code
update b set b.[sum]=b.[sum]+a.ssum
from tb b
(select id,sum(value) as ssum from ta group by id)a on a.id=b.id

insert into tb
select id,sum(value) from ta
  where not exists(select 1 from tb where id=ta.id) 
    group by id

------解决方案--------------------
declare @T table(id int,v int)
insert into @T
select 1,34 union all
select 2,56 union all  
select 6,18 union all  
select 1,42 union all  
select 5,274 union all  
select 1,56 union all  
select 2,56 

  
declare @A table(id int,v int)
select id,sum(v) as sumV from @T group by id

insert into @A  select id,sum(v) as sumV from @T group by id
select * from @A