Question

我的表格中有一个日期时间列，我想更新它，以便它可以获得今天的日期，但保留时间部分。

示例：

2011-01-21 01:12...  -> 2013-04-23 01:12...
2012-04-20 19:22...  -> 2013-04-23 19:22...

最简单的方法是什么？

关于奥斯卡

Answer 1

试试这个 -

DECLARE @date_old DATETIME
SELECT @date_old = '2011-01-21 01:12'

DECLARE @date_new DATETIME
SELECT @date_new = '2013-04-23 05:24'

SELECT CAST(CAST(@date_new AS DATE) AS DATETIME) + CAST(@date_old AS TIME)

Answer 2

修改以下是经过测试的解决方案

DECLARE @Today DATETIME2(7)
Declare @OldDate DateTime2(7)
SET @Today=GETDATE()

Select @OldDate = date From <SomeDatabase>

SELECT DATEADD(day, -DATEDIFF(day, @Today, @OldDate), @OldDate)

这是更新：

Update <SomeDatabase>  Set date = DATEADD(day, -DATEDIFF(day, @Today, @OldDate), @OldDate) where date = @OldDate

Answer 3

要更新表，请使用self join并使用@Devart的方法保留time，但更新date值

Declare @Sample table ( myDate datetime )
Insert into @Sample
values
('2013-04-21 11:42:51.897'),('2013-04-22 13:42:51.897')
Select * from @Sample

Update t
set myDate =convert(datetime,convert(date,getdate())) + convert (time,s.mydate)
from @Sample t inner join @Sample s
on s.myDate=t.myDate
Select * from @Sample

初步结果

╔═════════════════════════╗
║         myDate          ║
╠═════════════════════════╣
║ 2013-04-21 11:42:51.897 ║
║ 2013-04-22 13:42:51.897 ║
╚═════════════════════════╝

最终结果

╔═════════════════════════╗
║         myDate          ║
╠═════════════════════════╣
║ 2013-04-23 11:42:51.897 ║
║ 2013-04-23 13:42:51.897 ║
╚═════════════════════════╝

来自两个不同日期时间的时间和日期

3 个答案: