表：
deptID datatime money
1 2006-03-10 100
2 2006-03-10 200
3 2006-03-10 300
4 2006-03-10 400
1 2007-02-10 1000
2 2007-02-10 2000
3 2007-02-10 3000
4 2007-02-10 4000
结果： 1 2 3 4 合计
月份 2006 2007 2006 2007 2006 2007 2006 2007 2006 2007
02 0 1000 0 2000 0 3000 0 4000 0 10000
03 100 0 200 0 300 0 400 0 1000 0
不知道能不能实现啊 。。。哎~~~
------解决方案--------------------
declare @a table(deptID int,datatime char(10),money int)
insert into @a select 1 , '2006-03-10 ' , 100
union all select 2 , '2006-03-10 ' , 200
union all select 3 , '2006-03-10 ' , 300
union all select 4 , '2006-03-10 ' , 400
union all select 1 , '2007-02-10 ' , 1000
union all select 2 , '2007-02-10 ' , 2000
union all select 3 , '2007-02-10 ' , 3000
union all select 4 , '2007-02-10 ' , 4000
select distinct substring(datatime,6,2)as yuefen,
sum(case when deptid=1 and left(datatime,4)= '2006 'then money else 0 end)a,
sum(case when deptid=1 and left(datatime,4)= '2007 'then money else 0 end)b,
sum(case when deptid=2 and left(datatime,4)= '2006 'then money else 0 end)c,
sum(case when deptid=2 and left(datatime,4)= '2007 'then money else 0 end)d,
sum(case when deptid=3 and left(datatime,4)= '2006 'then money else 0 end)e,
sum(case when deptid=3 and left(datatime,4)= '2007 'then money else 0 end)f,