--写个最笨，但是最好理解的写法
declare @d datetime set @d = '2011-10-03'
declare @i int set @i=0

while(@d<getdate())
begin
    set @d=dateadd(month,1,@d)
    set @i=@i+1
end

if(datediff(d,getdate(),@d)>15)
set @i=@i-1

select @i as age
/*
age
-----------
4
*/

------解决方案--------------------
SQL code
DECLARE @D DateTime
SET @D = '2011-01-16'
select DATEDIFF(mm,@D,GETDATE())-1+  --月份掐头去尾
(case when DATEDIFF(d,@d,convert(varchar(8),dateadd(mm,1,@d),120)+'01')+DATEDIFF(d,convert(varchar(8),getdate(),120)+'01',GETDATE())>15 then 1 else 0 end)
/*
-----------
13

(1 行受影响)

*/

------解决方案--------------------
SQL code

--貌似这样就可以，楼主可以测试一下，有问题留言

declare @d datetime set @d= '2011-10-03'
declare @t datetime set @t= '2012-02-09'

select datediff(month,@d,getdate())+ceiling((day(@t)-day(@d))/15) as age
/*
age
-----------
4
*/

------解决方案--------------------
探讨


select *,case when datediff(dd,begindate,getdate())%30<=15 
then datediff(dd,begindate,getdate())/30 
else datediff(dd,begindate,getdate())/30+1 end
as [工龄(单位：月)] from emp


------解决方案--------------------
探讨

SQL code


--#9变量没放进来，修正一下
declare @d datetime set @d= '2011-10-03'
declare @t datetime set @t= '2012-02-09'

select datediff(month,@d,@t)+ceiling((day(@t)-day(@d))/15) as age
/*
age
-------……


------解决方案--------------------
SQL code

declare @d datetime set @d= '2011-02-28'
declare @t datetime set @t= '2011-03-30'

select case when day(@d)< 15 and day(@t)>=15   then datediff(mm,@d,@t)+ (case when abs(day(@d)-day(@t))>=15 then 1 else 0 end)
            when day(@d)< 15 and day(@t)< 15   then datediff(mm,@d,@t)
            when day(@d)>= 15 and day(@t)< 15  then datediff(mm,@d,@t)
            when day(@d)>= 15 and day(@t)>= 15 then datediff(mm,@d,@t)+ (case when abs(day(@d)-day(@t))>=15 then 1 else 0 end)
end as age 

试试这个