从SQL Server的DateTime字段获取'date'

Question

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

我只想让日期（而不是时间）超出该值。我怎么做？

Answer 1

如果您使用SQL Server 2008，现在有一个DATE数据类型。使它更自然！

SELECT CONVERT(Date, GETDATE()) 

Answer 2

这里：

SELECT creation_date 
FROM risks 
WHERE creation_date = GETDATE() 

这将返回存储在risks表是完全一样的东西是由GETDATE()函数返回的所有creation_date值。我假设数据类型creation_date是Date。

Answer 3

你只需要包括CREATION_DATE在你的SELECT子句是这样的：

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

Answer 4

如果我收到了你的问题的权利，

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

看CAST and CONVERT

Answer 5

它被称为“地板日期时间”，这样做是为了消除时间（这是最快的方法，比使用C更快） ONVERT（）或CAST（）蜇格式）：

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

OUTPUT：

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

(1 row(s) affected) 

这里是如何做到这一点的日期时间的其他部分：

--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 

OUTPUT：

-------- ----------------------- 
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 6

您可以随时使用月/日/年函数返回它：

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