9

我不知道它是否对某人有用，但我疯狂地寻找解决方案并最终自己做。这是一个函数（根据作为参数传递的日期），返回相同的日期和近似时间到最接近的 5 的倍数。这是一个慢查询，所以如果有人有更好的解决方案，欢迎。问候。

CREATE FUNCTION [dbo].[RoundTime] (@Time DATETIME) RETURNS DATETIME
AS
BEGIN
DECLARE @min nvarchar(50)
DECLARE @val int
DECLARE @hour int
DECLARE @temp int
DECLARE @day datetime
DECLARE @date datetime

SET @date = CONVERT(DATETIME, @Time, 120)

SET @day = (select DATEADD(dd, 0, DATEDIFF(dd, 0, @date)))
SET @hour = (select datepart(hour,@date))
SET @min = (select datepart(minute,@date))

IF LEN(@min) > 1
BEGIN
    SET @val = CAST(substring(@min, 2, 1) as int)
END
else
BEGIN
    SET @val = CAST(substring(@min, 1, 1) as int)
END

IF @val <= 2
BEGIN
    SET @val = CAST(CAST(@min as int) - @val as int)
END
else
BEGIN
    IF (@val <> 5)
    BEGIN
        SET @temp = 5 - CAST(@min%5 as int)
        SET @val = CAST(CAST(@min as int) + @temp as int)
    END

    IF (@val = 60)
    BEGIN
        SET @val = 0
        SET @hour = @hour + 1
    END

    IF (@hour = 24)
    BEGIN
        SET @day = DATEADD(day,1,@day)
        SET @hour = 0
        SET @min = 0
    END
END

RETURN CONVERT(datetime, CAST(DATEPART(YYYY, @day) as nvarchar) + '-' +  CAST(DATEPART(MM, @day) as nvarchar) + '-' + 
CAST(DATEPART(dd, @day) as nvarchar) + ' ' + CAST(@hour as nvarchar) + ':' +     CAST(@val as nvarchar), 120)

结尾

4

2 回答 2

20

您可以计算自纪元以来的分钟数：

datediff(minute,0,getdate())

通过除以整数 15 并乘以四舍五入到 15 分钟：

datediff(minute,0,getdate()) / 15 * 15

并将纪元添加回来：

dateadd(minute, datediff(minute,0,getdate()) / 15 * 15, 0)

例如：

select  dateadd(minute, datediff(minute,0,getdate()) / 15 * 15, 0)
-->
2013-11-12 10:45:00.000
于 2013-11-12T09:50:05.660 回答
15

创建这个函数

create function f_round5min
(
@date datetime
) returns datetime
as
begin -- adding 150 seconds to round off instead of truncating
return dateadd(minute, datediff(minute, '1900-01-01', dateadd(second, 150, @date))/5*5, 0)
end

使用此语法：

declare @testtable table(date datetime)

insert @testtable values('2013-11-12 12:00'),('2013-11-12 12:01'),
('2013-11-12 12:02'),('2013-11-12 12:02:29'),('2013-11-12 12:02:30'),
('2013-11-12 12:02:31'),('2013-11-12 12:03'),('2013-11-12 12:04'),
('2013-11-12 12:05')

select date, dbo.f_round5min(date) rounded from @testtable

结果：

date                     rounded
2013-11-12 12:00:00.000  2013-11-12 12:00:00.000
2013-11-12 12:01:00.000  2013-11-12 12:00:00.000
2013-11-12 12:02:00.000  2013-11-12 12:00:00.000
2013-11-12 12:02:29.000  2013-11-12 12:00:00.000
2013-11-12 12:02:30.000  2013-11-12 12:05:00.000
2013-11-12 12:02:31.000  2013-11-12 12:05:00.000
2013-11-12 12:03:00.000  2013-11-12 12:05:00.000
2013-11-12 12:04:00.000  2013-11-12 12:05:00.000
2013-11-12 12:05:00.000  2013-11-12 12:05:00.000
于 2013-11-12T09:53:54.823 回答