Question

表

id date-of-birth

001 01/01/2011 'dd/mm/yyyy'
002 05/01/2012
003 15/05/2009
....

从上表中，我想计算从加入日期列开始的天数，出生日期应该从当前年份进行验证。

查找日期差异查询

Select id, DATEDIFF(dd,Convert(datetime, date-of-join, 103),getdate())

以上查询是从加入日期开始的，但我想验证这样的加入日期......

例如

id date-of-birth no-of-days

001 01/01/2011   64 
002 05/01/2012   60
003 15/05/2009   295

...

条件

For 001, date-of-birth is '01/01/2011', so one year exceeded, then it should give no-0f-days from '01/01/2012'

For 002, date-of-birth is '05/01/2012, so it is not exceeded one years, then it should give no-of-days from '05/01/2012'

For 003, date-of-birth is '15/05/2009', so it is exceeded more than a years, then it should calculate from 15/05/2011 to current date

任何人都可以提供一些想法或查询帮助

Answer 1

尝试：

select id,
       [date-of-birth],
       datediff(yy,[date-of-birth],getdate()) -
                   case when dateadd(yy,datediff(yy,[date-of-birth],getdate()),[date-of-birth])>getdate()
                   then 1 else 0 end as [no-of-years],
       datediff(d,
                dateadd(yy,datediff(yy,[date-of-birth],getdate()) -
                           case when dateadd(yy,datediff(yy,[date-of-birth],getdate()),[date-of-birth])>getdate()
                           then 1 else 0 end,[date-of-birth]),
                getdate()) as [no-of-days]
from ...

Answer 2

declare @T table (id int identity(100,1), date datetime)
insert @T values('2012-05-01')
insert @T values('2012-03-1')
insert @T values('2011-01-01')
insert @T values('2009-05-15')
insert @T values('2008-01-23')

select 
    id,
    date,
    Days = case when datediff(dd,date,getdate())>730 
                    then datediff(dd,date,getdate()) - (datediff(dd,date,getdate())/365 * 365)
                when datediff(dd,date,getdate())>365 
                    then datediff(dd,convert(datetime, datename(yy,getdate()) + '/01/01'),getdate()) 
               else abs(datediff(dd,date,getdate()))
           end
from @T

如何找到确切的年份和日期

2 个答案: