如果你找到你的年龄。 Select (months_between(sysdate,dob)/12) from table_name 如果你找到你的大致年龄。 从table_name中选择round(months_between(sysdate,dob)/12);在这里输入代码 dob是列名。用于当前日期的Sysdate。Months_between用于查找dob到当前日期的总月份

闰年/日和以下方法有问题，请参阅下面的更新:

try this: DECLARE @dob datetime SET @dob='1992-01-09 00:00:00' SELECT DATEDIFF(hour,@dob,GETDATE())/8766.0 AS AgeYearsDecimal ,CONVERT(int,ROUND(DATEDIFF(hour,@dob,GETDATE())/8766.0,0)) AS AgeYearsIntRound ,DATEDIFF(hour,@dob,GETDATE())/8766 AS AgeYearsIntTrunc OUTPUT: AgeYearsDecimal AgeYearsIntRound AgeYearsIntTrunc --------------------------------------- ---------------- ---------------- 17.767054 18 17 (1 row(s) affected)

以下是一些更准确的方法:

多年来最好的方法

DECLARE @Now  datetime, @Dob datetime
SELECT   @Now='1990-05-05', @Dob='1980-05-05'  --results in 10
--SELECT @Now='1990-05-04', @Dob='1980-05-05'  --results in  9
--SELECT @Now='1989-05-06', @Dob='1980-05-05'  --results in  9
--SELECT @Now='1990-05-06', @Dob='1980-05-05'  --results in 10
--SELECT @Now='1990-12-06', @Dob='1980-05-05'  --results in 10
--SELECT @Now='1991-05-04', @Dob='1980-05-05'  --results in 10

SELECT
    (CONVERT(int,CONVERT(char(8),@Now,112))-CONVERT(char(8),@Dob,112))/10000 AS AgeIntYears

您可以将上面的10000更改为10000.0并获得小数，但它不会像下面的方法那样准确。

用十进制表示年份的最佳方法

DECLARE @Now  datetime, @Dob datetime
SELECT   @Now='1990-05-05', @Dob='1980-05-05' --results in 10.000000000000
--SELECT @Now='1990-05-04', @Dob='1980-05-05' --results in  9.997260273973
--SELECT @Now='1989-05-06', @Dob='1980-05-05' --results in  9.002739726027
--SELECT @Now='1990-05-06', @Dob='1980-05-05' --results in 10.002739726027
--SELECT @Now='1990-12-06', @Dob='1980-05-05' --results in 10.589041095890
--SELECT @Now='1991-05-04', @Dob='1980-05-05' --results in 10.997260273973

SELECT 1.0* DateDiff(yy,@Dob,@Now) 
    +CASE 
         WHEN @Now >= DATEFROMPARTS(DATEPART(yyyy,@Now),DATEPART(m,@Dob),DATEPART(d,@Dob)) THEN  --birthday has happened for the @now year, so add some portion onto the year difference
           (  1.0   --force automatic conversions from int to decimal
              * DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,@Now),DATEPART(m,@Dob),DATEPART(d,@Dob)),@Now) --number of days difference between the @Now year birthday and the @Now day
              / DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,@Now),1,1),DATEFROMPARTS(DATEPART(yyyy,@Now)+1,1,1)) --number of days in the @Now year
           )
         ELSE  --birthday has not been reached for the last year, so remove some portion of the year difference
           -1 --remove this fractional difference onto the age
           * (  -1.0   --force automatic conversions from int to decimal
                * DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,@Now),DATEPART(m,@Dob),DATEPART(d,@Dob)),@Now) --number of days difference between the @Now year birthday and the @Now day
                / DATEDIFF(day,DATEFROMPARTS(DATEPART(yyyy,@Now),1,1),DATEFROMPARTS(DATEPART(yyyy,@Now)+1,1,1)) --number of days in the @Now year
             )
     END AS AgeYearsDecimal

标记为正确的答案更接近准确，但在以下情况下，它失败了——出生年份是闰年，日在二月之后

declare @ReportStartDate datetime = CONVERT(datetime, '1/1/2014'),
@DateofBirth datetime = CONVERT(datetime, '2/29/1948')

FLOOR(DATEDIFF(HOUR,@DateofBirth,@ReportStartDate )/8766)

OR

FLOOR(DATEDIFF(HOUR,@DateofBirth,@ReportStartDate )/8765.82) -- Divisor is more accurate than 8766

下面的解决方案给了我更准确的结果。

FLOOR(DATEDIFF(YEAR,@DateofBirth,@ReportStartDate) - (CASE WHEN DATEADD(YY,DATEDIFF(YEAR,@DateofBirth,@ReportStartDate),@DateofBirth) > @ReportStartDate THEN 1 ELSE 0 END ))

它几乎适用于所有场景，包括闰年、2月29日等。

如果这个公式有漏洞，请指正。

CREATE function dbo.AgeAtDate(
    @DOB    datetime,
    @CompareDate datetime
)

returns INT
as
begin

return CASE WHEN @DOB is null
THEN 
    null
ELSE 
DateDiff(yy,@DOB, @CompareDate) 
- CASE WHEN datepart(mm,@CompareDate) > datepart(mm,@DOB) OR (datepart(mm,@CompareDate) = datepart(mm,@DOB) AND datepart(dd,@CompareDate) >= datepart(dd,@DOB))
  THEN 0 
  ELSE 1
  END
END
End

GO

所以上面的很多解决方案都是错误的DateDiff(yy，@Dob， @PassedDate)不会考虑两个日期的月和日。同样，只有在正确排序的情况下，省道部件才能进行比较。

下面的代码非常简单:

create function [dbo].[AgeAtDate](
    @DOB    datetime,
    @PassedDate datetime
)

returns int
with SCHEMABINDING
as
begin

declare @iMonthDayDob int
declare @iMonthDayPassedDate int


select @iMonthDayDob = CAST(datepart (mm,@DOB) * 100 + datepart  (dd,@DOB) AS int) 
select @iMonthDayPassedDate = CAST(datepart (mm,@PassedDate) * 100 + datepart  (dd,@PassedDate) AS int) 

return DateDiff(yy,@DOB, @PassedDate) 
- CASE WHEN @iMonthDayDob <= @iMonthDayPassedDate
  THEN 0 
  ELSE 1
  END

End

这将正确地处理生日和舍入的问题:

DECLARE @dob  datetime
SET @dob='1992-01-09 00:00:00'

SELECT DATEDIFF(YEAR, '0:0', getdate()-@dob)