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
以下是我如何计算年龄给出出生日期和当前日期。
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
其他回答
以下是我如何计算年龄给出出生日期和当前日期。
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
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
在尝试了许多方法后，这是100%的工作时间使用现代MS SQL格式函数，而不是转换为风格112。两者都可以，但这是最少的代码。
谁能找到一个日期组合不工作?我不认为有一个:)
--Set parameters, or choose from table.column instead:
DECLARE @DOB DATE = '2000/02/29' -- If @DOB is a leap day...
,@ToDate DATE = '2018/03/01' --...there birthday in this calculation will be
--0+ part tells SQL to calc the char(8) as numbers:
SELECT [Age] = (0+ FORMAT(@ToDate,'yyyyMMdd') - FORMAT(@DOB,'yyyyMMdd') ) /10000
这个怎么样:
SET @Age = CAST(DATEDIFF(Year, @DOB, @Stamp) as int)
IF (CAST(DATEDIFF(DAY, DATEADD(Year, @Age, @DOB), @Stamp) as int) < 0)
SET @Age = @Age - 1
SELECT ID,
Name,
DATEDIFF(yy,CONVERT(DATETIME, DOB),GETDATE()) AS AGE,
DOB
FROM MyTable
