在T-SQL（SQL Server 2000）中以年，月和日计算某人年龄的最佳方法是什么？
datediff函数不能很好地处理年份边界，加上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，但我想在我的stored procedure中完成它。
答案 0 :(得分：60)
这是一些T-SQL，它为您提供自@date中指定日期以来的年数，月数和天数。它考虑到DATEDIFF（）在不考虑它的月份或日期的情况下计算差异的事实（因此8/31和9/1之间的月份差异为1个月）并使用case语句处理该结果，该语句将结果递减到合适的。
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
答案 1 :(得分：10)
试试这个......
SELECT CASE WHEN
(DATEADD(year,DATEDIFF(year, @datestart ,@dateend) , @datestart) > @dateend)
THEN DATEDIFF(year, @datestart ,@dateend) -1
ELSE DATEDIFF(year, @datestart ,@dateend)
END
基本上是“DateDiff（年......”，给你这个人将在今年转的年龄，所以我只是添加一个案例陈述说，如果他们今年还没有过生日，那么减去1年，否则返回值。
答案 2 :(得分：8)
将文字称为年龄的简单方法如下：
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：
**63 Y & 2 M**
答案 3 :(得分：4)
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
日期的方法与接受的答案不同，差异显示在以下评论中：
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
set @days = CASE WHEN day(@now) >= day(@dob) THEN day(@now) - day(@dob)
ELSE @days_in_birth_month - day(@dob) + day(@now) END
如果您只想要年龄和月份，可能会更简单
set @years = ( @now_i/100 - @dob_i/100)/100
set @months =(12 + month(@now) - month(@dob))%12
select @years,@months -- 1, 0
答案 4 :(得分：3)
这是一个（稍微）更简单的版本：
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
答案 5 :(得分：3)
与功能相同的东西。
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
答案 6 :(得分：2)
答案 7 :(得分：1)
我已经多次看到这个问题，结果输出年，月，日但从不是数字/小数结果。 （至少没有一个不能圆不正确）。 我欢迎有关此功能的反馈。可能还不需要稍微调整一下。
- 该功能的输入是两个日期。 - 输出是十进制（7,4）格式的两个日期之间的数字年数。 - 输出始终是可能的数字。
- 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）差异恰好是整数年（1,2,3等）
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
`
答案 8 :(得分：1)
我使用我在@Dane答案：https://stackoverflow.com/a/57720/2097023
中修改的函数（“天数”部分）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。
答案 9 :(得分：1)
有一种简单的方法，根据两天之间的小时数而截断结束日期。
SELECT CAST(DATEDIFF(hour,Birthdate,CAST(GETDATE() as Date))/8766.0 as INT) AS Age FROM <YourTable>
这个已被证明是非常准确和可靠的。如果它不是GETDATE（）上的内部CAST，它可能会在午夜前几个小时翻转生日，但是，在CAST中，它已经死了，年龄在正好午夜时会改变。
答案 10 :(得分：1)
相当古老的问题，但我想分享我为计算年龄而做的事情
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)
答案 11 :(得分：0)
还有另一种计算年龄的方法是
请参见下表
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
Select datediff(MONTH,DOB,getdate())/12 as dates from [Organization].[Employee]
结果将是
firstname dates
sai 27
Harish 20
答案 12 :(得分：0)
您是否正在计算一个年龄的总天数/月/年？你有开始约会吗？或者你想解剖它（例如：24年，1个月，29天）？
如果您有一个与之合作的开始日期，则datediff将使用以下命令输出总天数/月/年：
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')
各自的输出为（8827/290/24）。
现在，如果你想要进行解剖方法，你必须减去以天为单位的年数（天数 - 365 *年），然后再进行数学计算以获得月份等等。
答案 13 :(得分：0)
以下是我如何计算出生日期和当前日期的年龄。
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
答案 14 :(得分：0)
对于想要在表中创建计算列以存储年龄的人：
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
答案 15 :(得分：0)
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.
答案 16 :(得分：0)
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')
答案 17 :(得分：0)
这是SQL代码，它为您提供自sysdate以来的年数，月数和天数。 输入input_birth_date这个格式的值（dd_mon_yy）。注意：输入相同的值（出生日期）年，月和日日期如01-mar-85
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
答案 18 :(得分：0)
DateTime值存储为浮点数。您可以相互减去日期，现在您有一个新的日期，即它们之间的时间跨度。
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)
答案 19 :(得分：-1)
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
答案 20 :(得分：-1)
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]
答案 21 :(得分：-1)
select DOB as Birthdate,
YEAR(GETDATE()) as ThisYear,
YEAR(getdate()) - EAR(date1) as Age
from TableName
答案 22 :(得分：-1)
SELECT DOB AS Birthdate ,
YEAR(GETDATE()) AS ThisYear,
YEAR(getdate()) - YEAR(DOB) AS Age
FROM tableprincejain