declare @dt datetime

set @dt = '09-22-2007 15:07:38.850'

select dateadd(mi, datediff(mi, 0, @dt), 0)
select dateadd(hour, datediff(hour, 0, @dt), 0)

会回来

2007-09-22 15:07:00.000
2007-09-22 15:00:00.000

以上只是截断了秒和分钟，产生了问题中要求的结果.正如@OMG Ponies 指出的，如果你想向上/向下舍入，那么你可以分别添加半分钟或半小时，然后截断:

The above just truncates the seconds and minutes, producing the results asked for in the question. As @OMG Ponies pointed out, if you want to round up/down, then you can add half a minute or half an hour respectively, then truncate:

select dateadd(mi, datediff(mi, 0, dateadd(s, 30, @dt)), 0)
select dateadd(hour, datediff(hour, 0, dateadd(mi, 30, @dt)), 0)

你会得到:

2007-09-22 15:08:00.000
2007-09-22 15:00:00.000

在 SQL Server 2008 中添加 date 数据类型之前，我会使用上面的从日期时间截断时间部分以仅获取日期的方法.这个想法是确定有问题的日期时间和固定时间点之间的天数(0，它隐式转换为 1900-01-01 00:00:00.000>):

Before the date data type was added in SQL Server 2008, I would use the above method to truncate the time portion from a datetime to get only the date. The idea is to determine the number of days between the datetime in question and a fixed point in time (0, which implicitly casts to 1900-01-01 00:00:00.000):

declare @days int
set @days = datediff(day, 0, @dt)

然后将该天数添加到固定时间点，这将为您提供原始日期，时间设置为 00:00:00.000:

and then add that number of days to the fixed point in time, which gives you the original date with the time set to 00:00:00.000:

select dateadd(day, @days, 0)

或更简洁:

select dateadd(day, datediff(day, 0, @dt), 0)

使用不同的日期部分(例如 hour、mi)会相应地起作用.

Using a different datepart (e.g. hour, mi) will work accordingly.