关于tsql:如何用年,月和日来计算T-SQL中的年龄

关于tsql:如何用年,月和日来计算T-SQL中的年龄

How to calculate age in T-SQL with years, months, and days

在T-SQL(SQL Server 2000)中计算某人的年,月和日的年龄的最佳方法是什么?

datediff函数不能很好地处理年份界限,另外,将月份和日期分开将是一个负担。 我知道我可以在客户端相对容易地做到这一点,但是我想在我的存储过程中完成它。


这是一些T-SQL,可为您提供@date中指定日期以来的年数,月数和天数。它考虑到DATEDIFF()计算差异而没有考虑是哪个月或一天的事实(因此8/31和9/1之间的月份差异为1个月),并使用case语句处理该差异,从而减少结果,其中适当。

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DECLARE @date datetime, @tmpdate datetime, @years int, @months int, @days int
SELECT @date = '2/29/04'

SELECT @tmpdate = @date

SELECT @years = DATEDIFF(yy, @tmpdate, GETDATE()) - CASE WHEN (MONTH(@date) > MONTH(GETDATE())) OR (MONTH(@date) = MONTH(GETDATE()) AND DAY(@date) > DAY(GETDATE())) THEN 1 ELSE 0 END
SELECT @tmpdate = DATEADD(yy, @years, @tmpdate)
SELECT @months = DATEDIFF(m, @tmpdate, GETDATE()) - CASE WHEN DAY(@date) > DAY(GETDATE()) THEN 1 ELSE 0 END
SELECT @tmpdate = DATEADD(m, @months, @tmpdate)
SELECT @days = DATEDIFF(d, @tmpdate, GETDATE())

SELECT @years, @months, @days

尝试这个...

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SELECT CASE WHEN
 (DATEADD(year,DATEDIFF(year, @datestart  ,@dateend) , @datestart) > @dateend)
THEN DATEDIFF(year, @datestart  ,@dateend) -1
ELSE DATEDIFF(year, @datestart  ,@dateend)
END

基本上," DateDiff(year ...")为您提供了该人今年的年龄,因此,我仅添加一个案例说明,如果他们今年还没有生日,则减去1年,否则返回值。


通过文本获取年龄的简单方法如下:

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Select cast((DATEDIFF(m, date_of_birth, GETDATE())/12) as varchar) + ' Y & ' +
       cast((DATEDIFF(m, date_of_birth, GETDATE())%12) as varchar) + ' M' as Age

结果格式为:

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**63 Y & 2 M**

通过算术实现,采用ISO格式的日期。

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declare @now date,@dob date, @now_i int,@dob_i int, @days_in_birth_month int
declare @years int, @months int, @days int
set @now = '2013-02-28'
set @dob = '2012-02-29' -- Date of Birth

set @now_i = convert(varchar(8),@now,112) -- iso formatted: 20130228
set @dob_i = convert(varchar(8),@dob,112) -- iso formatted: 20120229
set @years = ( @now_i - @dob_i)/10000
-- (20130228 - 20120229)/10000 = 0 years

set @months =(1200 + (month(@now)- month(@dob))*100 + day(@now) - day(@dob))/100 %12
-- (1200 + 0228 - 0229)/100 % 12 = 11 months

set @days_in_birth_month = day(dateadd(d,-1,left(convert(varchar(8),dateadd(m,1,@dob),112),6)+'01'))
set @days = (sign(day(@now) - day(@dob))+1)/2 * (day(@now) - day(@dob))
          + (sign(day(@dob) - day(@now))+1)/2 * (@days_in_birth_month - day(@dob) + day(@now))
-- ( (-1+1)/2*(28 - 29) + (1+1)/2*(29 - 29 + 28))
-- Explain: if the days of now is bigger than the days of birth, then diff the two days
--          else add the days of now and the distance from the date of birth to the end of the birth month
select @years,@months,@days -- 0, 11, 28

测试用例

天数的方法与接受的答案不同,以下注释中显示了差异:

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       dob        now  years  months  days
2012-02-29 2013-02-28      0      11    28  --Days will be 30 if calculated by the approach in accepted answer.
2012-02-29 2016-02-28      3      11    28  --Days will be 31 if calculated by the approach in accepted answer, since the day of birth will be changed to 28 from 29 after dateadd by years.
2012-02-29 2016-03-31      4       1     2
2012-01-30 2016-02-29      4       0    30
2012-01-30 2016-03-01      4       1     2  --Days will be 1 if calculated by the approach in accepted answer, since the day of birth will be changed to 30 from 29 after dateadd by years.
2011-12-30 2016-02-29      4       1    30

Days by case语句的简短版本:

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set @days = CASE WHEN day(@now) >= day(@dob) THEN day(@now) - day(@dob)
                 ELSE @days_in_birth_month - day(@dob) + day(@now) END

如果您只需要几岁和几个月的年龄,可能会更简单

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set @years = ( @now_i/100 - @dob_i/100)/100
set @months =(12 + month(@now) - month(@dob))%12
select @years,@months -- 1, 0

注意:SQL Server日期格式的一个非常有用的链接


这是一个(略)简单的版本:

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CREATE PROCEDURE dbo.CalculateAge
    @dayOfBirth datetime
AS

DECLARE @today datetime, @thisYearBirthDay datetime
DECLARE @years int, @months int, @days int

SELECT @today = GETDATE()

SELECT @thisYearBirthDay = DATEADD(year, DATEDIFF(year, @dayOfBirth, @today), @dayOfBirth)

SELECT @years = DATEDIFF(year, @dayOfBirth, @today) - (CASE WHEN @thisYearBirthDay > @today THEN 1 ELSE 0 END)

SELECT @months = MONTH(@today - @thisYearBirthDay) - 1

SELECT @days = DAY(@today - @thisYearBirthDay) - 1

SELECT @years, @months, @days
GO

与功能相同的事物。

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create function [dbo].[Age](@dayOfBirth datetime, @today datetime)
   RETURNS varchar(100)
AS

Begin
DECLARE @thisYearBirthDay datetime
DECLARE @years int, @months int, @days int

set @thisYearBirthDay = DATEADD(year, DATEDIFF(year, @dayOfBirth, @today), @dayOfBirth)
set @years = DATEDIFF(year, @dayOfBirth, @today) - (CASE WHEN @thisYearBirthDay > @today THEN 1 ELSE 0 END)
set @months = MONTH(@today - @thisYearBirthDay) - 1
set @days = DAY(@today - @thisYearBirthDay) - 1

return cast(@years as varchar(2)) + ' years,' + cast(@months as varchar(2)) + ' months,' + cast(@days as varchar(3)) + ' days'
end

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create  procedure getDatedifference

(
    @startdate datetime,
    @enddate datetime
)
as
begin
    declare @monthToShow int
    declare @dayToShow int

    --set @startdate='01/21/1934'
    --set @enddate=getdate()

    if (DAY(@startdate) > DAY(@enddate))
        begin
            set @dayToShow=0

            if (month(@startdate) > month(@enddate))
                begin
                    set @monthToShow=  (12-month(@startdate)+ month(@enddate)-1)
                end
            else if (month(@startdate) < month(@enddate))
                begin
                    set @monthToShow=  ((month(@enddate)-month(@startdate))-1)
                end
            else
               begin
                   set @monthToShow=  11
               end
            -- set @monthToShow= convert(int, DATEDIFF(mm,0,DATEADD(dd,DATEDIFF(dd,0,@enddate)- DATEDIFF(dd,0,@startdate),0)))-((convert(int,FLOOR(DATEDIFF(day, @startdate, @enddate) / 365.25))*12))-1
                         if(@monthToShow<0)
                         begin
                            set @monthToShow=0
                         end

                      declare @amonthbefore integer
                      set @amonthbefore=Month(@enddate)-1
                          if(@amonthbefore=0)
                             begin
                                set @amonthbefore=12
                              end


                      if (@amonthbefore  in(1,3,5,7,8,10,12))
                          begin
                            set @dayToShow=31-DAY(@startdate)+DAY(@enddate)
                          end
                      if (@amonthbefore=2)
                         begin
                           IF (YEAR( @enddate ) % 4 = 0 AND YEAR( @enddate ) % 100 != 0) OR  YEAR( @enddate ) % 400 = 0
                                 begin
                                    set @dayToShow=29-DAY(@startdate)+DAY(@enddate)
                                  end
                           else
                               begin
                                   set @dayToShow=28-DAY(@startdate)+DAY(@enddate)
                           end
                      end
                      if (@amonthbefore in (4,6,9,11))
                        begin
                           set @dayToShow=30-DAY(@startdate)+DAY(@enddate)
                        end
                 end
    else
        begin
          --set @monthToShow=convert(int, DATEDIFF(mm,0,DATEADD(dd,DATEDIFF(dd,0,@enddate)- DATEDIFF(dd,0,@startdate),0)))-((convert(int,FLOOR(DATEDIFF(day, @startdate, @enddate) / 365.25))*12))
          if (month(@enddate)< month(@startdate))
              begin
                 set @monthToShow=12+(month(@enddate)-month(@startdate))
              end
          else
              begin
                set @monthToShow= (month(@enddate)-month(@startdate))
              end
          set @dayToShow=DAY(@enddate)-DAY(@startdate)
        end

    SELECT
        FLOOR(DATEDIFF(day, @startdate, @enddate) / 365.25) as [yearToShow],
          @monthToShow as  monthToShow ,@dayToShow as dayToShow ,
        convert(varchar,FLOOR(DATEDIFF(day, @startdate, @enddate) / 365.25)) +' Year ' + convert(varchar,@monthToShow) +' months '+convert(varchar,@dayToShow)+' days ' as age

    return
end

我已经看过几次这个问题了,结果输出的是年,月,日,但是从来没有数字/小数的结果。 (至少没有一个不会四舍五入的错误)??。
欢迎对此功能提出反馈。可能仍不需要一点调整。

-该函数的输入是两个日期。
-输出是以Decimal(7,4)格式显示的两个日期之间的年数。
-输出始终是一个正数。

-注意:如果差异大于999.9999,则输出将无法处理

-逻辑基于三个步骤。
-1)差异是否小于1年(0.5000、0.3333、0.6667等)
-2)差异是否是整数年(1、2、3等)?

-3)(其他)...不同之处在于年份和天数。 (1.5000、2.3333、7.6667等)

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CREATE Function [dbo].[F_Get_Actual_Age](@pi_date1 datetime,@pi_date2 datetime)
RETURNS Numeric(7,4)
AS
BEGIN

Declare
 @l_tmp_date    DATETIME
,@l_days1       DECIMAL(9,6)
,@l_days2       DECIMAL(9,6)
,@l_result      DECIMAL(10,6)
,@l_years       DECIMAL(7,4)


  --Check to make sure there is a date for both inputs
  IF @pi_date1 IS NOT NULL and @pi_date2 IS NOT NULL  
  BEGIN

    IF @pi_date1 > @pi_date2 --Make sure the"older" date is in @pi_date1
      BEGIN
        SET @l_tmp_date = @pi_date2
        SET @pi_date2 = @Pi_date1
        SET @pi_date1 = @l_tmp_date
      END

    --Check #1 If date1 + 1 year is greater than date2, difference must be less than 1 year
    IF DATEADD(YYYY,1,@pi_date1) > @pi_date2  
      BEGIN
          --How many days between the two dates (numerator)
        SET @l_days1 = DATEDIFF(dd,@pi_date1, @pi_date2)
          --subtract 1 year from date2 and calculate days bewteen it and date2
          --This is to get the denominator and accounts for leap year (365 or 366 days)
        SET @l_days2 = DATEDIFF(dd,dateadd(yyyy,-1,@pi_date2),@pi_date2)
        SET @l_years = @l_days1 / @l_days2 -- Do the math
      END
    ELSE
      --Check #2  Are the dates an exact number of years apart.
      --Calculate years bewteen date1 and date2, then add the years to date1, compare dates to see if exactly the same.
      IF DATEADD(YYYY,DATEDIFF(YYYY,@pi_date1,@pi_date2),@pi_date1) = @pi_date2  
        SET @l_years = DATEDIFF(YYYY,@pi_date1, @pi_date2) --AS Years, 'Exactly even Years' AS Msg
      ELSE
      BEGIN
        --Check #3 The rest of the cases.
        --Check if datediff, returning years, over or under states the years difference
        SET @l_years = DATEDIFF(YYYY,@pi_date1, @pi_date2)
        IF DATEADD(YYYY,@l_years,@pi_date1) > @pi_date2
          SET @l_years = @l_years -1
          --use basicly same logic as in check #1  
        SET @l_days1 = DATEDIFF(dd,DATEADD(YYYY,@l_years,@pi_date1), @pi_date2)
        SET @l_days2 = DATEDIFF(dd,dateadd(yyyy,-1,@pi_date2),@pi_date2)
        SET @l_years = @l_years + @l_days1 / @l_days2
        --SELECT @l_years AS Years, 'Years Plus' AS Msg
      END
  END
  ELSE
    SET @l_years = 0  --If either date was null

RETURN @l_Years  --Return the result as decimal(7,4)
END

`


相当老的问题,但我想分享我在计算年龄时所做的工作

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    Declare @BirthDate As DateTime
Set @BirthDate = '1994-11-02'

SELECT DATEDIFF(YEAR,@BirthDate,GETDATE()) - (CASE
WHEN MONTH(@BirthDate)> MONTH(GETDATE()) THEN 1
WHEN MONTH(@BirthDate)= MONTH(GETDATE()) AND DAY(@BirthDate) > DAY(GETDATE()) THEN 1
Else 0 END)


这是一种简单的方法,基于BUT两天之间的小时数,且结束日期被截断。

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SELECT CAST(DATEDIFF(hour,Birthdate,CAST(GETDATE() as Date))/8766.0 as INT) AS Age FROM <YourTable>

这已被证明是非常准确和可靠的。如果不是在GETDATE()上使用内部CAST,则可能会在午夜之前几个小时翻转生日,但是使用CAST,随着年龄恰好在午夜改变,它就死定了。


我使用从@Dane答复中修改的此功能("天"部分):https://stackoverflow.com/a/57720/2097023

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CREATE FUNCTION dbo.EdadAMD
    (
        @FECHA DATETIME
    )
    RETURNS NVARCHAR(10)
    AS
    BEGIN
        DECLARE
            @tmpdate DATETIME
          , @years   INT
          , @months  INT
          , @days    INT
          , @EdadAMD NVARCHAR(10);

        SELECT @tmpdate = @FECHA;

        SELECT @years = DATEDIFF(yy, @tmpdate, GETDATE()) - CASE
                                          WHEN (MONTH(@FECHA) >    MONTH(GETDATE()))
                                             OR (
                                                MONTH(@FECHA) = MONTH(GETDATE())
                                          AND DAY(@FECHA) > DAY(GETDATE())
                                          ) THEN
                                                1
                                            ELSE
                                                0
                                    END;
    SELECT @tmpdate = DATEADD(yy, @years, @tmpdate);
    SELECT @months = DATEDIFF(m, @tmpdate, GETDATE()) - CASE
                              WHEN DAY(@FECHA) > DAY(GETDATE()) THEN
                                                            1
                                                        ELSE
                                                            0
                                                    END;
    SELECT @tmpdate = DATEADD(m, @months, @tmpdate);

    IF MONTH(@FECHA) = MONTH(GETDATE())
       AND DAY(@FECHA) > DAY(GETDATE())
          SELECT @days =
            DAY(EOMONTH(GETDATE(), -1)) - (DAY(@FECHA) - DAY(GETDATE()));
    ELSE
        SELECT @days = DATEDIFF(d, @tmpdate, GETDATE());

    SELECT @EdadAMD = CONCAT(@years, 'a', @months, 'm', @days, 'd');

    RETURN @EdadAMD;

END;
GO

效果很好。


对于那些想要在表中创建计算列以存储年龄的用户:

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CASE WHEN DateOfBirth< DATEADD(YEAR, (DATEPART(YEAR, GETDATE()) - DATEPART(YEAR, DateOfBirth))*-1, GETDATE())
     THEN DATEPART(YEAR, GETDATE()) - DATEPART(YEAR, DateOfBirth)
     ELSE DATEPART(YEAR, GETDATE()) - DATEPART(YEAR, DateOfBirth) -1 END

还有另一种计算年龄的方法是

见下表

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    FirstName       LastName    DOB
    sai             krishnan    1991-11-04
    Harish          S A         1998-10-11

要查找年龄,您可以计算出一个月

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  Select datediff(MONTH,DOB,getdate())/12 as dates from [Organization].[Employee]

结果将是

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firstname   dates
sai         27
Harish      20

这是给定生日和当前日期的年龄计算方法。

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select case
            when cast(getdate() as date) = cast(dateadd(year, (datediff(year, '1996-09-09', getdate())), '1996-09-09') as date)
                then dateDiff(yyyy,'1996-09-09',dateadd(year, 0, getdate()))
            else dateDiff(yyyy,'1996-09-09',dateadd(year, -1, getdate()))
        end as MemberAge
go


您是否要计算年龄的总天数/月数/年数?您有开始日期吗?还是要剖析(例如:24年零1个月29天)?

如果您使用的是开始日期,datediff将使用以下命令输出总的天/月/年:

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Select DateDiff(d,'1984-07-12','2008-09-11')

Select DateDiff(m,'1984-07-12','2008-09-11')

Select DateDiff(yyyy,'1984-07-12','2008-09-11')

输出分别为(8827/290/24)。

现在,如果要使用解剖方法,则必须减去以天为单位的年数(天-365 *年),然后对其进行进一步的数学运算以得出月数,依此类推。


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CREATE FUNCTION DBO.GET_AGE
(
@DATE AS DATETIME
)
RETURNS VARCHAR(MAX)
AS
BEGIN

DECLARE @YEAR  AS VARCHAR(50) = ''
DECLARE @MONTH AS VARCHAR(50) = ''
DECLARE @DAYS  AS VARCHAR(50) = ''
DECLARE @RESULT AS VARCHAR(MAX) = ''

SET @YEAR  = CONVERT(VARCHAR,(SELECT DATEDIFF(MONTH,CASE WHEN DAY(@DATE) > DAY(GETDATE()) THEN DATEADD(MONTH,1,@DATE) ELSE @DATE END,GETDATE()) / 12 ))
SET @MONTH = CONVERT(VARCHAR,(SELECT DATEDIFF(MONTH,CASE WHEN DAY(@DATE) > DAY(GETDATE()) THEN DATEADD(MONTH,1,@DATE) ELSE @DATE END,GETDATE()) % 12 ))
SET @DAYS = DATEDIFF(DD,DATEADD(MM,CONVERT(INT,CONVERT(INT,@YEAR)*12 + CONVERT(INT,@MONTH)),@DATE),GETDATE())

SET @RESULT = (RIGHT('00' + @YEAR, 2) + ' YEARS ' + RIGHT('00' + @MONTH, 2) + ' MONTHS ' + RIGHT('00' + @DAYS, 2) + ' DAYS')

RETURN @RESULT
END

SELECT DBO.GET_AGE('04/12/1986')

这是SQL代码,为您提供自sysdate以来的年数,月数和天数。
输入此格式的输入日期(date_birth_date)的值(dd_mon_yy)。注意:输入年,月,日的相同值(出生日期),例如85年3月1日

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select trunc((sysdate -to_date('&input_birth_date_dd_mon_yy'))/365) years,
trunc(mod(( sysdate -to_date('&input_birth_date_dd_mon_yy'))/365,1)*12) months,
trunc((mod((mod((sysdate -to_date('&input_birth_date_dd_mon_yy'))/365,1)*12),1)*30)+1) days
 from dual

T-SQL中的DateTime值存储为浮点数。您可以相互减去日期,现在有了一个新的日期,即它们之间的时间跨度。

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declare @birthdate datetime
set @birthdate = '6/15/1974'

--age in years - short version
print year(getdate() - @birthdate) - year(0)

--age in years - visualization
declare @mindate datetime
declare @span datetime

set @mindate = 0
set @span = getdate() - @birthdate

print @mindate
print @birthdate
print getdate()
print @span
--substract minyear from spanyear to get age in years
print year(@span) - year(@mindate)
print month(@span)
print day(@span)

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DECLARE @BirthDate datetime, @AgeInMonths int
SET @BirthDate = '10/5/1971'
SET @AgeInMonths                              -- Determine the age in"months old":
    = DATEDIFF(MONTH, @BirthDate, GETDATE())  -- .Get the difference in months
    - CASE WHEN DATEPART(DAY,GETDATE())       -- .If today was the 1st to 4th,
              < DATEPART(DAY,@BirthDate)      --   (or before the birth day of month)
           THEN 1 ELSE 0 END                  --   ... don't count the month.
SELECT @AgeInMonths / 12 as AgeYrs            -- Divide by 12 months to get the age in years
      ,@AgeInMonths % 12 as AgeXtraMonths     -- Get the remainder of dividing by 12 months = extra months
      ,DATEDIFF(DAY                           -- For the extra days, find the difference between,
               ,DATEADD(MONTH, @AgeInMonths   -- 1. Last Monthly Birthday
                             , @BirthDate)    --     (if birthdays were celebrated monthly)
               ,GETDATE()) as AgeXtraDays     -- 2. Today's date.


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DECLARE @DoB AS DATE = '1968-10-24'
DECLARE @cDate AS DATE = CAST('2000-10-23' AS DATE)

SELECT
--Get Year difference
DATEDIFF(YEAR,@DoB,@cDate) -
--Cases where year difference will be augmented
CASE
    --If Date of Birth greater than date passed return 0
    WHEN YEAR(@DoB) - YEAR(@cDate) >= 0 THEN DATEDIFF(YEAR,@DoB,@cDate)

    --If date of birth month less than date passed subtract one year
    WHEN MONTH(@DoB) - MONTH(@cDate) > 0 THEN 1

    --If date of birth day less than date passed subtract one year
    WHEN MONTH(@DoB) - MONTH(@cDate) = 0 AND DAY(@DoB) - DAY(@cDate) > 0 THEN 1

    --All cases passed subtract zero
    ELSE 0
END

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declare @StartDate datetime = '2016-01-31'
declare @EndDate datetime = '2016-02-01'
SELECT @StartDate AS [StartDate]
      ,@EndDate AS [EndDate]
      ,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END AS [Years]
      ,DATEDIFF(Month,(DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate)),@EndDate) - CASE WHEN DATEADD(Month, DATEDIFF(Month,DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate),@EndDate) , @StartDate) > @EndDate THEN 1 ELSE 0 END AS [Months]
      ,DATEDIFF(Day, DATEADD(Month,DATEDIFF(Month, (DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate)),@EndDate) - CASE WHEN DATEADD(Month, DATEDIFF(Month,DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate),@EndDate) , @StartDate) > @EndDate THEN 1 ELSE 0 END  ,DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate)) ,@EndDate) - CASE WHEN DATEADD(Day,DATEDIFF(Day, DATEADD(Month,DATEDIFF(Month, (DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate)),@EndDate) - CASE WHEN DATEADD(Month, DATEDIFF(Month,DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate),@EndDate) , @StartDate) > @EndDate THEN 1 ELSE 0 END  ,DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate)) ,@EndDate),DATEADD(Month,DATEDIFF(Month, (DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate)),@EndDate) - CASE WHEN DATEADD(Month, DATEDIFF(Month,DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate),@EndDate) , @StartDate) > @EndDate THEN 1 ELSE 0 END  ,DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate) - CASE WHEN DATEADD(Year,DATEDIFF(Year,@StartDate,@EndDate), @StartDate) > @EndDate THEN 1 ELSE 0 END,@StartDate))) > @EndDate THEN 1 ELSE 0 END AS [Days]

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select DOB as Birthdate,
       YEAR(GETDATE()) as ThisYear,
       YEAR(getdate()) - EAR(date1) as Age  
from TableName

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SELECT DOB AS Birthdate ,
       YEAR(GETDATE()) AS ThisYear,
       YEAR(getdate()) - YEAR(DOB) AS Age
FROM tableprincejain


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