How to calculate age (in years) based on Date of Birth and getDate()(如何根据出生日期和 getDate() 计算年龄(以年为单位))
问题描述
我有一张表格,列出了人们的出生日期(目前是 nvarchar(25))
I have a table listing people along with their date of birth (currently a 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)?
How can I convert that to a date, and then calculate their age in years?
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:
ID Name AGE DOB
1 John 17 1992-01-09 00:00:00
2 Sally 50 1959-05-20 00:00:00
推荐答案
闰年/天有问题,方法如下,看下面更新:
There are issues with leap year/days and the following method, see the update below:
试试这个:
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
输出:
AgeYearsDecimal AgeYearsIntRound AgeYearsIntTrunc
--------------------------------------- ---------------- ----------------
17.767054 18 17
(1 row(s) affected)
更新这里有一些更准确的方法:
UPDATE here are some more accurate methods:
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
你可以把上面的10000
改成10000.0
并得到小数,但不会像下面的方法那么准确.
you can change the above 10000
to 10000.0
and get decimals, but it will not be as accurate as the method below.
多年来最好的十进制方法
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
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本文标题为:如何根据出生日期和 getDate() 计算年龄(以年为单位)
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