Question

我有一个日期列，其中日期以2009-11-18 10:55:28.370格式显示。

我只想从该值中获取日期（而不是时间）。我该怎么做？

Answer 1

如果您使用的是SQL Server 2008，那么现在有一种DATE数据类型。让它更自然！

SELECT CONVERT(Date, GETDATE())

Answer 2

它被称为“地板日期时间”，这样做只是为了删除时间（这是最快的方法，比使用CONVERT（）或CAST（）sting格式化更快）：

DECLARE @datetime datetime;
SET @datetime = '2008-09-17 12:56:53.430';
SELECT DATEADD(day,DATEDIFF(day,0,@datetime),0)

输出：

-----------------------
2008-09-17 00:00:00.000

(1 row(s) affected)

以下是如何为日期时间的其他部分执行此操作：

--Floor a datetime
DECLARE @datetime datetime;
SET @datetime = '2008-09-17 12:56:53.430';

SELECT '0 None',  @datetime                                                                   -- none    2008-09-17 12:56:53.430
UNION SELECT '1 Second',DATEADD(second,DATEDIFF(second,'2000-01-01',@datetime),'2000-01-01')  -- Second: 2008-09-17 12:56:53.000
UNION SELECT '2 Minute',DATEADD(minute,DATEDIFF(minute,0,@datetime),0)                        -- Minute: 2008-09-17 12:56:00.000
UNION SELECT '3 Hour',  DATEADD(hour,DATEDIFF(hour,0,@datetime),0)                            -- Hour:   2008-09-17 12:00:00.000
UNION SELECT '4 Day',   DATEADD(day,DATEDIFF(day,0,@datetime),0)                              -- Day:    2008-09-17 00:00:00.000
UNION SELECT '5 Month', DATEADD(month,DATEDIFF(month,0,@datetime),0)                          -- Month:  2008-09-01 00:00:00.000
UNION SELECT '6 Year',  DATEADD(year,DATEDIFF(year,0,@datetime),0)                            -- Year:   2008-01-01 00:00:00.000
ORDER BY 1
PRINT' '
PRINT 'Note that when you are flooring by the second, you will often get an arithmetic overflow if you use 0. So pick a known value that is guaranteed to be lower than the datetime you are attempting to floor'
PRINT 'this always uses a date less than the given date, so there will be no arithmetic overflow'
SELECT '1 Second',DATEADD(second,DATEDIFF(second,DATEADD(day,DATEDIFF(day,0,@datetime),0)-1,@datetime),DATEADD(day,DATEDIFF(day,0,@datetime),0)-1)  -- Second: 2008-09-17 12:56:53.000

输出：

-------- -----------------------
0 None   2008-09-17 12:56:53.430
1 Second 2008-09-17 12:56:53.000
2 Minute 2008-09-17 12:56:00.000
3 Hour   2008-09-17 12:00:00.000
4 Day    2008-09-17 00:00:00.000
5 Month  2008-09-01 00:00:00.000
6 Year   2008-01-01 00:00:00.000

(7 row(s) affected)


Note that when you are flooring by the second, you will often get an arithmetic overflow if you use 0. So pick a known value that is guaranteed to be lower than the datetime you are attempting to floor
this always uses a date less than the given date, so there will be no arithmetic overflow

-------- -----------------------
1 Second 2008-09-17 12:56:53.000

(1 row(s) affected)

Answer 3

如果我的问题是对的，那么

select convert(varchar, creation_date , 103) as creation_date from tablename

查看CAST and CONVERT

Answer 4

下面：

SELECT creation_date 
FROM risks
WHERE creation_date = GETDATE()

这将返回creation_date表中存储的所有risks值，这些值与GETDATE()函数返回的值完全相同。我假设creation_date的数据类型为Date。

Answer 5

您只需要在您的select子句中包含creation_date，如下所示：

select id, creation_date from risks where creation_date = getdate()

Answer 6

您始终可以使用月/日/年功能将其退回：

declare @date datetime
set @date = '1/1/10 12:00 PM'
select cast(month(@date) as varchar) + '/' + cast(day(@date) as varchar) + '/' + cast(year(@date) as varchar) as theDate

从SQL Server中的DateTime字段中获取“日期”

6 个答案: