How can I make an algorithm that can give me the right values for year, month,day ,h,min,s?

I'm trying to make an algorithm that takes count if a year is a leap year and also to make it recognize the number of days of every month so that we have for example 31 december 2001 and the time is 23:59:00 and we add a minute we will obtain 1 jan 2002 00:00:00 . But the script seems to be faulty and the results aren't right.This is an algorithm for a function but I also want to put it in guide where a user could write the date using an editbox. The problem in this situation is what happens if a user puts as day 44 for example, The algorithm only takes account if the day is greater than 30, 31 ,28,29 and makes the day 1. Also another problem is if a user puts hour =0 , in my function I have that hour is 0 , day=day+1.THis condition was put because I have a tspan and the algorithm should work based the transformation of second in min,h,day.. Can somebody offer me some guidance over this problems that I'm trying to avoid? For the problem with the user input I was thinking of a notice message or something that won't let the user put a bigger value for that month.

 t=3600;
year=2001;
anb=0; %leap year
%condition for leap year
 if mod(year,4)~=0
        anb=0;
    else if mod(anb,100)~=0
            anb=1;
        else if mod(anb,400)~=0
                anb=0;
            else anb=1;
            end
        end
 end
month= 12;
day=28;
hour=23;
min=25;
sec=17;
sec = sec + rem(t, 60);
min = rem(min + floor(t/60), 60);
hour = rem(hour + floor(t/3600), 24);
%conditions
if hour==0
    day=day+1;
end

 if (anb==1 && month==2 && day >29)
                month=3;
                day=1;

            else if anb==0 && month==2 && day>28
                    month=3;
                    day=1;

                end
 end
if month==1||3||5||7||8||10 && day>31
    month=month+1
    day=1
else if month==4||6||9||11 && day>30
        month=month+1
        day=1
    else if month==12 && day>31
            year=year+1
            month=1
            day=1
        end
    end
end

Answer 1

[Steven Lord]

If this is NOT a homework assignment and you're using release R2014b or later, just represent your date and time data using datetime. [If this IS a homework assignment, I'm guessing your professor wants you to implement the date and time handling yourself.]

[Alexandru Lapusneanu]

Not an homework , just working for my graduation thesis. Thought it would look better if I had written the code by myself. But your suggestion is also a good one.

[Steven Lord]

BTW this doesn't do what you think it does.

if month==1||3||5||7||8||10 && day>31

If month is a scalar, this is equivalent to:

if day > 31

or

if (month==1) || 3 || 5 || 7 || 8 || 10 && day>31

The expression (anything) || 3 is true because 3 is nonzero. The expression true || 5 is true because both true and 5 are nonzero. Continuing that reasoning, it boils down to (true || 10) && (day > 31) or true && (day > 31).

If you want to check if a number is in a set of numbers, I recommend using the ismember or ismembertol functions.

[Edited to fix the missing or operators.]

[Alexandru Lapusneanu]

Thank you!