Round time to 5 minute nearest SQL Server

17,322

Solution 1

Create this function

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

Use this syntax:

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

Result:

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

Solution 2

You can calculate the number of minutes since the epoch:

datediff(minute,0,getdate())

Round to 15 minutes by dividing and multiplying by integer 15:

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

And add the epoch back:

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

For example:

select  dateadd(minute, datediff(minute,0,getdate()) / 15 * 15, 0)
-->
2013-11-12 10:45:00.000
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Drako
Author by

Drako

Updated on July 21, 2022

Comments

  • Drako
    Drako almost 2 years

    i don't know if it can be usefull to somebody but I went crazy looking for a solution and ended up doing it myself. Here is a function that (according to a date passed as parameter), returns the same date and approximate time to the nearest multiple of 5. It is a slow query, so if anyone has a better solution, it is welcome. A greeting.

    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)
    

    END

  • Morvael
    Morvael about 8 years
    Might be stating the obvious but if you actually want the time rounded up, just replace dateadd(second,150,@date) with just @date
  • Orin Moyer
    Orin Moyer almost 7 years
    Best non-function answear
  • Grandizer
    Grandizer almost 3 years
    It is the ONLY non-function answer. However, the above function by @t-clausen.dk could be written inline as SELECT DATEADD(MINUTE, DATEDIFF(MINUTE, '1900-01-01', DATEADD(SECOND, 150, MyDateField)) / 5 * 5, 0)