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 @FromDate DATETIME = '1992-01-2623:59:59.000',
@ToDate DATETIME = '2016-08-10 00:00:00.000',
@Years INT, @Months INT, @Days INT, @tmpFromDate DATETIME
SET @Years = DATEDIFF(YEAR, @FromDate, @ToDate)
- (CASE WHEN DATEADD(YEAR, DATEDIFF(YEAR, @FromDate, @ToDate),
@FromDate) > @ToDate THEN 1 ELSE 0 END)
SET @tmpFromDate = DATEADD(YEAR, @Years , @FromDate)
SET @Months = DATEDIFF(MONTH, @tmpFromDate, @ToDate)
- (CASE WHEN DATEADD(MONTH,DATEDIFF(MONTH, @tmpFromDate, @ToDate),
@tmpFromDate) > @ToDate THEN 1 ELSE 0 END)
SET @tmpFromDate = DATEADD(MONTH, @Months , @tmpFromDate)
SET @Days = DATEDIFF(DAY, @tmpFromDate, @ToDate)
- (CASE WHEN DATEADD(DAY, DATEDIFF(DAY, @tmpFromDate, @ToDate),
@tmpFromDate) > @ToDate THEN 1 ELSE 0 END)
SELECT @FromDate FromDate, @ToDate ToDate,
@Years Years, @Months Months, @Days Days
其他回答
DECLARE @yourBirthDate DATETIME = '1987-05-25'
SELECT YEAR(DATEADD(DAY, DATEDIFF(DAY, @yourBirthDate, GETDATE()), CAST('0001-01-01' AS DATETIME2))) - 1
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
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
下面的脚本检查现在和给定出生日期之间的年差;第二部分检查该生日在当年是否已经过去;如果不是，则减去:
SELECT year(NOW()) - year(date_of_birth) - (CONCAT(year(NOW()), '-', month(date_of_birth), '-', day(date_of_birth)) > NOW()) AS Age
FROM tableName;
select DATEDIFF(yy,@DATE,GETDATE()) -
case when DATEPART(mm,GETDATE())*100+DATEPART(dd,GETDATE())>=
DATEPART(mm,@DATE)*100+DATEPART(dd,@DATE) THEN 0
ELSE 1 END
