determine if a datetime value is on a different day

If I have two datetime values, t1 and t2, how can I determine if t2 is on a different, and later, day than t1? The check needs to account for end-of-month and end-of-year rollovers, and the times (hr, min, sec) may not be the same between t1 and t2. I just need to know if they are are different calendar days.

Functions like between and caldiff seem to return a duration, but if that duration is less than 24 hours, this method won't work. For example, two times only 2 minutes apart but accoss a day boundary:

t1 = datetime('2025-03-01 23:59:00')

t1 = datetime
   01-Mar-2025 23:59:00

t2 = datetime('2025-03-02 00:01:00')

t2 = datetime
   02-Mar-2025 00:01:00

dt = caldiff([t2 t1],'days')

dt = calendarDuration
   0d

between(t1,t2,'days')

ans = calendarDuration
   0d

datenum would work, but MATLAB is recommending not to use. How can this be done with datetime values and functions?

Answer 1

[Stephen23]

"Functions like between and caldiff seem to return a duration"

BETWEEN and CALDIFF will not help you. Either use DATESHIFT or the properties of the DATETIME object:

t1 = datetime('2025-03-01 23:59:00')

t1 = datetime
   01-Mar-2025 23:59:00

t2 = datetime('2025-03-02 00:01:00')

t2 = datetime
   02-Mar-2025 00:01:00

dateshift(t1,'start','day')==dateshift(t2,'start','day')

ans = logical
   0

t1.Year==t2.Year && t1.Month==t2.Month && t1.Day==t2.Day

ans = logical
   0

[Tim]

Ah, dateshift. That's a new one. I'm still learning all of the datetime functions. Thanks!

[Peter Perkins]

There's also this, if you are already doing something else with the time-of-day:

[T,D] = timeofday(DT) also returns the date portions of the values in DT as the datetime array D.

The output argument D is equivalent to dateshift(DT,'start','day').

dt = datetime;
[~,day] = timeofday(dt)

day = datetime
   03-Mar-2025

Answer 2

[Walter Roberson]

t1 = datetime('2025-03-01 23:59:00')

t1 = datetime
   01-Mar-2025 23:59:00

t2 = datetime('2025-03-02 00:01:00')

t2 = datetime
   02-Mar-2025 00:01:00

t2 >= t1 & dateshift(t1, 'start', 'day') == dateshift(t2, 'start', 'day')

ans = logical
   0