select a.uid,sum(b.money) from table2 a left join table1 b
gruop by a.uid

------解决方案--------------------
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

------解决方案--------------------
select sum([money]),uid summoney  from  table1 where uid in  
(select uid from table2)
group by table1.uid

------解决方案--------------------
SQL code

select sum([money]),a.uid summoney from table1  as a
join table2 as b on a.uid =b.uid 
group by a.uid

------解决方案--------------------
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   ……


------解决方案--------------------
SELECT DISTINCT uid,SUM(money)OVER(PARTITION BY uid ) FROM 有一张表记录消费信息的table1

------解决方案--------------------
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)
……


------解决方案--------------------
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(……


------解决方案--------------------
SQL code

create table #T 
(
[ID] int identity(1,1) NOT NULL,
[uid]int NULL,
[money] decimal(12,2) NULL
)  
INSERT INTO #T
SELECT  2, 3 union all
SELECT  2 ,5 union all
SELECT   1, 5 union all
SELECT   8, 3 union all
SELECT   5, 10 union all
SELECT  3, 20 union all
SELECT   8, 6 union all
SELECT   6, 9 union all
SELECT   8, 3 union all
SELECT   1, 60 union all
SELECT   3, 8 union all
SELECT   3, 5 union all
SELECT   5, 6 union all
SELECT  2, 10 union all
SELECT  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           65.00
2           18.00
3           43.00
5           16.00
6           9.00
8           12.00

(6 行受影响)

------解决方案--------------------
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 #T
SELECT  2, 3 union all
SELECT  2 ,5 union all
SELECT   1, 5 union all
SELECT   8, 3 union all
SELECT   5, 10 union all
SELECT  3, 20 union all
SELECT   8, 6 union all
SELECT   6, 9 union all
SELECT   8, 3 union all
SELECT   1, 60 union all
SELECT   3, 8 union all
SELECT   3, 5 union all
SELECT   5, 6 union all
SELECT  2, 10 union all
SELECT  3, 10 
go 

select uid,sum(m