sqlserver!遇到难题了

sqlserver高手请进!遇到难题了！
有一张表记录消费信息的table1
id uid money
1 2 3
2 2 5
3 1 5
4 8 3
5 5 10
6 3 20
7 8 6
8 6 9
9 8 3
10 1 60
11 3 8
12 3 5
13 5 6
14 2 10
15 3 10

现在想查出每一个uid 所消费的money.
uid 是从表table2是查询所得。

也就是说，从table2中查询不重复uid 去对应表table1中的uid，再查出每个uid的总消费(money)情况

求sql语句

速度解决，速度给分！



[解决办法]

SQL code

select a.uid,sum(b.money) from table2 a left join table1 bgruop by a.uid
[解决办法]
不怎么清晰的感觉。。。
table1的uid是有重复的  那到时候算总消费依据什么呢
额  或许我理解能力不够高啊
[解决办法]
select sum([money]),uid summoney  from  table1 where uid in  
(select uid from table2)
group by table1.uid
[解决办法]
SQL code
select sum([money]),a.uid summoney from table1  as ajoin table2 as b on a.uid =b.uid group by a.uid
[解决办法]
探讨


id uid money
1 2 3
2 2 5
3 1 5
4 8 3
5 5 10
6 3 20
7 8 6
8 6 9
9 8 3
10 1 60
11 3 8
12 3 5
13 5 6
14 2 10
15 3 10

我想得到的结果是

uid    money
 2    sum(money)
 1    sum(money)
 8   ……

[解决办法]
SELECT DISTINCT uid,SUM(money)OVER(PARTITION BY uid ) FROM 有一张表记录消费信息的table1
[解决办法]
探讨

id uid money
1 2 3
2 2 5
3 1 5
4 8 3
5 5 10
6 3 20
7 8 6
8 6 9
9 8 3
10 1 60
11 3 8
12 3 5
13 5 6
14 2 10
15 3 10

我想得到的结果是

uid money
 2 sum(money)
 1 sum(money)
 8 sum(money)
……

[解决办法]
探讨


引用:

引用:

id uid money
1 2 3
2 2 5
3 1 5
4 8 3
5 5 10
6 3 20
7 8 6
8 6 9
9 8 3
10 1 60
11 3 8
12 3 5
13 5 6
14 2 10
15 3 10

我想得到的结果是

uid    money
2    sum(……

[解决办法]
SQL code
create table #T ([ID] int identity(1,1) NOT NULL,[uid]int NULL,[money] decimal(12,2) NULL)  INSERT INTO #TSELECT  2, 3 union allSELECT  2 ,5 union allSELECT   1, 5 union allSELECT   8, 3 union allSELECT   5, 10 union allSELECT  3, 20 union allSELECT   8, 6 union allSELECT   6, 9 union allSELECT   8, 3 union allSELECT   1, 60 union allSELECT   3, 8 union allSELECT   3, 5 union allSELECT   5, 6 union allSELECT  2, 10 union allSELECT  3, 10 SELECT DISTINCT uid,SUM(money)OVER(PARTITION BY uid ) Ssum FROM #t drop table #T(15 行受影响)uid         Ssum----------- ---------------------------------------1           65.002           18.003           43.005           16.006           9.008           12.00(6 行受影响)
[解决办法]
貌似这个快很多
SQL code
create table #T ([ID] int identity(1,1) NOT NULL,[uid]int NULL,[money] decimal(12,2) NULL)  INSERT INTO #TSELECT  2, 3 union allSELECT  2 ,5 union allSELECT   1, 5 union allSELECT   8, 3 union allSELECT   5, 10 union allSELECT  3, 20 union allSELECT   8, 6 union allSELECT   6, 9 union allSELECT   8, 3 union allSELECT   1, 60 union allSELECT   3, 8 union allSELECT   3, 5 union allSELECT   5, 6 union allSELECT  2, 10 union allSELECT  3, 10 go select uid,sum(money)Ssum from #t group by uid drop table #T 
 
[解决办法]
考虑使用视图+函数？

create function fn_getMoney(@id int)
 RETURNS decimal(18,4)
AS
 BEGIN
  declare @re decimal(18,4);

   select @re = sum(money)
  from table1
  where id = @id
  return ISNULL(@re , 0);
 END
GO

;with c1 as
(
 select distinct uid from tb
)
select c1.uid, fn_getMoney(c1.uid) as SumMoney
from c1  

不晓得利用函数来进行直接筛选，效果会不会好一点，呵呵，验证下……