求一存储过程。根据出生年月日得出年龄，要求精确到岁，月，日
1981-01-01   得到   26岁
2006-12-01   得到   1月
2007-01-01   得到   14天
要求返回年龄和年龄单位   ,26岁,1月,14天
寻求最简单的写法

------解决方案--------------------
declare @dt datetime
set @dt= '1981-01-01 '
select
(
datediff(year, @dt, getdate())-
case when dateadd(year, datediff(year, @dt, getdate()), @dt)> getdate()
then 1 else 0 end
) as 年龄,
datediff(month, dateadd( year, datediff(year, @dt, getdate()), @dt ), getdate()) as 月,
datediff(day, dateadd(month, datediff(month, @dt, getdate()), @dt), getdate()) as 天

--result
年龄 月 天
----------- ----------- -----------
26 0 14

(1 row(s) affected)
------解决方案--------------------
还是不对~
----
declare @dt datetime
set @dt= '1981-08-20 '
select
(
datediff(year, @dt, getdate())-
case when dateadd(year, datediff(year, @dt, getdate()), @dt)> getdate()
then 1 else 0 end
) as 年龄,
abs( datediff(month, dateadd( year, datediff(year, @dt, getdate()), @dt ), getdate()) ) as 月,
abs( datediff(day, dateadd(month, datediff(month, @dt, getdate()), @dt), getdate()) ) as 天

----
25 7 5

------解决方案--------------------
declare @dt datetime, @year int, @month int
set @dt= '1981-01-01 '

select @year = datediff(year, @dt, getdate()), @month = datediff(month, @dt, getdate())

if(@year > 0 and dateadd(year, @year, @dt) <=getdate())
begin
select cast(@year as varchar(10))+ '岁 '
end else
begin
if(@month > 1)
begin
if(dateadd(month, @month, @dt)> getdate())
begin
set @month = @month - 1
end

select cast(@month as varchar(10))+ '月 '
end else
if(@month > 0 and dateadd(month, @month, @dt) <=getdate())
begin
select cast(@month as varchar(10))+ '月 '
end else
begin
select cast(datediff(day, @dt, getdate()) as varchar(10))+ '天 '
end
end