我有一个表格,其中列出了人员及其出生日期(当前为 nvarchar(25))
如何将其转换为日期,然后计算他们的年龄(以年为单位)?
我的数据如下
ID Name DOB
1 John 1992-01-09 00:00:00
2 Sally 1959-05-20 00:00:00
我想看看:
ID Name AGE DOB
1 John 17 1992-01-09 00:00:00
2 Sally 50 1959-05-20 00:00:00
最佳答案
闰年/闰日和以下方法存在问题,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)
更新这里有一些更准确的方法:
INT 多年来的最佳方法
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
你可以将上面的10000
改为10000.0
并得到小数,但它不会像下面的方法那么准确。
多年来十进制的最佳方法
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
关于sql - 如何根据出生日期和 getDate() 计算年龄(以岁为单位),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33427165/