在线等：求一条SQL语句！！
问题是这样的：
一个表 fy（费用表），字段有：fyje（费用金额）,dxje (兑现金额) ，wlh（往来户）,fyrq（费用日期）,dxrq（兑现日期）
同一个往来户（wlh） 有N条费用。
例如：
wlh, fyje, dxje ， fyrq, dxrq
泛洋 250 250 2007-05-01 2007-09-12
泛洋 100 100 2007-09-11 2007-09-12
泛洋 150 150 2007-09-12 2007-10-25
泛洋 200 200 2007-09-15 2007-10-25
只要dxrq 大于fyrq 表示费用已经兑现了 fyje就是 兑现金额
现在要一条这样的语句：
求往来户的所有金额，1<(dxrq-fyrq）<30天的兑现金额，（30<(dxrq-fyrq)<60的兑现金额,60<(fxrq-fyrq）的金额



------解决方案--------------------
乱写的，可能结果不对，但是我觉得应该可以这样写
------解决方案--------------------
这样写就ok了

select distinct Sum(dxje) as sumtotal, 
(select Sum(dxje) 
from fy 
where datediff(dd, fyrq, dxrq) between 1 and 30 ) as '1-30',
(select Sum(dxje) 
from fy 
where datediff(dd, fyrq, dxrq) between 31 and 60 ) as '31-60' ,
(select Sum(dxje) 
from fy 
where datediff(dd, fyrq, dxrq) >60 ) as '>60'
 from fy 

------解决方案--------------------
晕 应该这样 
select distinct wlh,Sum(fyje) as sumtotal,
(select Sum(dxje)
from fy
where wlh = a.wlh and datediff(dd, fyrq, dxrq) between 1 and 30) as '1-30 ',
(select Sum(dxje)
from fy
where wlh = a.wlh and datediff(dd, fyrq, dxrq) between 31 and 60) as '31-60 ',
(select Sum(dxje)
from fy
where wlh = a.wlh and datediff(dd, fyrq, dxrq) >60) as ' >60 ' 
from fy a 
group by Wlh
------解决方案--------------------
select wlh,
sum(case when datediff(dd,fyrq,dxrq)<=1 then 0 when datediff(dd,fyrq,dxrq)>=30 then o else fyje end) as '30天内的兑现金额',
sum(case when datediff(dd,fyrq,dxrq)<=30 then 0 when datediff(dd,fyrq,dxrq)>=60 then o else fyje end) as '30天到60天内的兑现金额',
sum(case when datediff(dd,fyrq,dxrq)>=60 then 0 else fyje end) as '60天内的兑现金额'
from ty
group by wlh
order by wlh