SQL DateDiff在几分钟内跨越整个夜晚

Question

Trying to get the difference in minutes between a datetime field and a time field. I'm using the datediff function.

When the start time starts on one date like '2018-01-08 22:35:55.043' and the end time is the following day like '00:35:56.2136644', the result is counting from the end time to the start time.

Examples:

select DATEDIFF(MINUTE, CAST('2018-01-08 22:35:55.043' AS TIME), '00:35:56.2136644') AS minDiff1
select DATEDIFF(MINUTE,  '00:35:56.2136644', CAST('2018-01-08 22:35:55.043' AS TIME)) AS minDiff2
select DATEDIFF(MINUTE, CONVERT(TIME, '2018-01-08 22:35:55.043'), '00:35:56.2136644') AS minDiff3
select DATEDIFF(MINUTE, '00:35:56.2136644', CONVERT(TIME, '2018-01-08 22:35:55.043')) AS minDiff4

The results were different from what I was expecting. The desired result would be 120 minutes.

minDiff1 = -1320

minDiff2 = 1320

minDiff3 = -1320

minDiff4 = 1320

Original Query

select DATEDIFF(MINUTE, CAST(test_start_datetime as TIME), test_end_time) AS minDiff
from user_exam  

Answer 1

The following assumes that the time belongs to the same day or the very next day:

SELECT *, CASE
    -- same day -- start time is less than end time
    WHEN CAST(datetimecol AS time) <= timecol THEN DATEDIFF(MINUTE, CAST(datetimecol AS time), timecol)
    -- next day -- start time is more than end time (it rolled over into next day)
    ELSE 1440 - DATEDIFF(MINUTE, timecol, CAST(datetimecol AS time))
END
FROM (VALUES
    (CAST('2018-01-08 22:35:55.043' AS DATETIME), CAST('22:35:55.0433333' AS TIME)),
    (CAST('2018-01-08 22:35:55.043' AS DATETIME), CAST('23:35:56.2136644' AS TIME)),
    (CAST('2018-01-08 22:35:55.043' AS DATETIME), CAST('00:35:56.2136644' AS TIME))
) AS tests(datetimecol, timecol)

In the above example 1440 is the number of minutes in 24 hours.

Demo on DB Fiddle

Answer 2

Let us first reconsider your examples:

select DATEDIFF(MINUTE, CAST('2018-01-08 22:35:55.043' AS TIME), '00:35:56.2136644') AS minDiff1
select DATEDIFF(MINUTE,  '00:35:56.2136644', CAST('2018-01-08 22:35:55.043' AS TIME)) AS minDiff2
select DATEDIFF(MINUTE, CONVERT(TIME, '2018-01-08 22:35:55.043'), '00:35:56.2136644') AS minDiff3
select DATEDIFF(MINUTE, '00:35:56.2136644', CONVERT(TIME, '2018-01-08 22:35:55.043')) AS minDiff4

Converting/casting from datetime to time throws the date part away. Thus, you are actually running:

select DATEDIFF(MINUTE, '22:35:55.043'', '00:35:56.2136644') AS minDiff1
select DATEDIFF(MINUTE,  '00:35:56.2136644', '22:35:55.043') AS minDiff2
select DATEDIFF(MINUTE, '22:35:55.043', '00:35:56.2136644') AS minDiff3
select DATEDIFF(MINUTE, '00:35:56.2136644'22:35:55.043') AS minDiff4

At this point, you seem to not be considering the fact that DATEDIFF is directed, meaning it counts (first argument units passed) FROM second argument TO third argument. Thus, since 00:35 is 1320 minutes earlier than 22:35 (of the same day), 00:35 -> 22:35 returns 1320, whereas 22:35 -> 00:35 returns -1320.

---

To be completely precise, since DATEDIFF uses datetimes, your times/time-representing strings are implicitly converted to dates. Since a date is not provided by you, the date used is the one with value 0: 1st January 1900. That's the common day the function acts upon.

Answer 3

If you want to see 120 that's mean time difference is not enough for your purpose. With your question you want to find difference between 22:35 and following day 00:35 You need to find DATETIME difference something like that:

SELECT DATEDIFF(MINUTE,         CAST(GETDATE() AS DATETIME)
                               +CAST(CAST('22:35:55.2136644' AS TIME) AS DATETIME)
                             , 
                                CAST(GETDATE() AS DATETIME)+1
                               +CAST(CAST('00:35:56.2136644' AS TIME)AS DATETIME)

                ) AS minDiff1