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例如表的结构，表之间的关系，测试数据，相关算法及需要的结果。
这样有助于我们理解你的意思，更主要的是能尽快让你获得答案或解决问题的方法。

------解决方案--------------------
SQL code
declare @sql varchar(8000)
set @sql = 'select isnull(userid,''合计'') as 合计'
select @sql = @sql + ' , sum(case convert(varchar(10),time,120) when ''' + convert(varchar(10),time,120) + ''' then price else 0 end) [' + convert(varchar(10),time,120) + ']'
from (select distinct convert(varchar(10),time,120) from tb) as a
set @sql = @sql + ' from tb group by userid with rollup'
exec(@sql)

------解决方案--------------------
datediff(month,[date],getdate())=0筛选本月，做行转列。

------解决方案--------------------

SQL code
create table pay(--支出
id int identity(1,1) primary key,--主键
pay_title varchar(100),--支出名称
pay_name varchar(100),--支出人姓名
type_id int ,--foreign key references paytype(id),--支出类型（外键）
price float,--支出金额
time date,--支出时间
brief varchar(1000),--支出简介
userid int --foreign key references users(id)--支出人Id
)
insert into pay(pay_title,[time],price) select 'a','2011-07-01',542.21
insert into pay(pay_title,[time],price) select 'a','2011-07-01',985.25
insert into pay(pay_title,[time],price) select 'a','2011-07-02',246.55
insert into pay(pay_title,[time],price) select 'a','2011-07-03',81849.55
insert into pay(pay_title,[time],price) select 'a','2011-07-04',57154.21
insert into pay(pay_title,[time],price) select 'a','2011-08-04',74.21
go
select time,price into # from pay where CONVERT(varchar(7),[time],120)='2011-07'
declare @s nvarchar(4000)
select @s=isnull(@s+',','')+'['+ convert(varchar(10),time,120) +']' from 
(select distinct time from #)t
exec('select '+@s+'from # pivot (sum([price]) for [time] in('+@s+'))b')
/*
2011-07-01             2011-07-02             2011-07-03             2011-07-04
---------------------- ---------------------- ---------------------- ----------------------
1527.46                246.55                 81849.55               57154.21

(1 行受影响)

*/
go
drop table pay,#
go