我有一个表,上面列出了人们的出生日期(目前是nvarchar(25))

我如何将其转换为日期,然后以年为单位计算他们的年龄?

我的数据如下所示

ID    Name   DOB
1     John   1992-01-09 00:00:00
2     Sally  1959-05-20 00:00:00

我希望看到:

ID    Name   AGE  DOB
1     John   17   1992-01-09 00:00:00
2     Sally  50   1959-05-20 00:00:00

当前回答

Declare @dob datetime
Declare @today datetime

Set @dob = '05/20/2000'
set @today = getdate()

select  CASE
            WHEN dateadd(year, datediff (year, @dob, @today), @dob) > @today 
            THEN datediff (year, @dob, @today) - 1
            ELSE datediff (year, @dob, @today)
        END as Age

其他回答

试试这个解决方案:

declare @BirthDate datetime
declare @ToDate datetime

set @BirthDate = '1/3/1990'
set @ToDate = '1/2/2008'
select @BirthDate [Date of Birth], @ToDate [ToDate],(case when (DatePart(mm,@ToDate) <  Datepart(mm,@BirthDate)) 
        OR (DatePart(m,@ToDate) = Datepart(m,@BirthDate) AND DatePart(dd,@ToDate) < Datepart(dd,@BirthDate))
        then (Datepart(yy, @ToDate) - Datepart(yy, @BirthDate) - 1)
        else (Datepart(yy, @ToDate) - Datepart(yy, @BirthDate))end) Age

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)

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
select floor((datediff(day,0,@today) - datediff(day,0,@birthdate)) / 365.2425) as age

这里有很多365.25的答案。记住闰年是如何定义的:

每四年一次 除了每100年 除了每400年

我已经在我们的生产代码中使用这个查询近10年了:

SELECT FLOOR((CAST (GetDate() AS INTEGER) - CAST(Date_of_birth AS INTEGER)) / 365.25) AS Age