Select 查询问题

[kyle2cj]

Month fee
20090112 10
20090123 20
20090203 15
20090227 16
20090301 11

怎么样写查询语句得到：按Month对Fee求和

Month fee
200901 30
200902 31
200903 11

[csdyyr]

DECLARE @TB TABLE(Month VARCHAR(8),          fee INT)
INSERT @TB
SELECT '20090112',  10 UNION ALL 
SELECT '20090123',  20 UNION ALL 
SELECT '20090203',  15 UNION ALL 
SELECT '20090227',  16 UNION ALL 
SELECT '20090301',  11


SELECT LEFT(MONTH,6) AS [MONTH],SUM(FEE) AS FEE
FROM @TB 
GROUP BY LEFT(MONTH,6)
/*
MONTH    FEE         
-------- ----------- 
200901   30
200902   31
200903   11
*/

[dawugui]

select left([month],6) [month] , sum(fee) fee from tb group by left([month],6)

[jia_guijun]

select left(month,6) as Month,sum(fee) as fee from tb group by left(month,6)

[dawugui]

create table tb([Month] varchar(10), fee int)
insert into tb values('20090112' , 10 )
insert into tb values('20090123' , 20 )
insert into tb values('20090203' , 15 )
insert into tb values('20090227' , 16 )
insert into tb values('20090301' , 11 )
go

select left([month],6) [month] , sum(fee) fee from tb group by left([month],6)

drop table tb

/*
month      fee         
---------- ----------- 
200901     30
200902     31
200903     11

（所影响的行数为 3 行）

*/

[nj_1st_excellence]

select substring(Month,1,6) Month, sum(fee) fee from tb group by substring(Month,1,6)

[dawugui]

--如果month为datetime型.

create table tb([Month] datetime, fee int)
insert into tb values('20090112' , 10 )
insert into tb values('20090123' , 20 )
insert into tb values('20090203' , 15 )
insert into tb values('20090227' , 16 )
insert into tb values('20090301' , 11 )
go

select convert(varchar(6),[month],112) [month], sum(fee) fee from tb group by convert(varchar(6),[month],112)

drop table tb

/*
month      fee         
---------- ----------- 
200901     30
200902     31
200903     11

（所影响的行数为 3 行）

*/

[htl258_Tony]

[Quote=引用楼主 kyle2cj 的帖子:]
Month          fee 
20090112      10 
20090123      20 
20090203      15 
20090227      16 
20090301      11 

怎么样写查询语句得到：按Month对Fee求和 

Month        fee 
200901        30 
200902        31 
200903        11
[/Quote]
create table tb([Month] datetime, fee int)
insert into tb values('20090112' , 10 )
insert into tb values('20090123' , 20 )
insert into tb values('20090203' , 15 )
insert into tb values('20090227' , 16 )
insert into tb values('20090301' , 11 )
go

select convert(varchar(6),[month],112) [Month],sum(fee) fee 
from tb 
group by convert(varchar(6),[month],112)
order by [Month]

drop table tb

Month  fee
------ -----------
200901 30
200902 31
200903 11

(3 行受影响)

[kyle2cj]

没怎么写过查询，这么简单，也没想到，，，谢谢各位
结贴，给分