Calculate day number from an unix-timestamp in a math way?

Question

How can i calculate day number from a unix-timestamp, in a mathematical way and without using any functions and in simple math formula.

1313905026 --> 8 (Today 08/21

Answer 1

t = unix time
second = t MOD 60  
minute = INT(t / 60) MOD 60  
hour = INT(t / 60 / 60) MOD 24  
days = INT(t / 60 / 60 / 24)  
years = INT(days / 365.25)  
year = 1970 + years + 1

1970 started with a Thursday so, we can calculate the day of the week:

weekday = (days + 4) MOD 7

If Sunday is day 0. If you want Sunday to be day 1 just add 1.

Now, let's find out how many days we are into the year in question.

days = days - years * 365 - leapdays

Finally, we find the month and day of the month.

IF year MOD 4 = 0 THEN ly = 1 ELSE ly = 0
WHILE month <= 12
    month = month + 1
    IF month = 2 THEN
        DaysInMonth = 28 + NOT(year MOD 4) + NOT(year MOD 100)
            + NOT(year MOD 400)
    ELSE
        DaysInMonth = 30 + (month + (month < 7)) MOD 2
    END IF
    IF days > DaysInMonth THEN days = days - DaysInMonth
END WHILE

This assumes Boolean values of TRUE = 1, FALSE = 0, NOT TRUE = 0, and NOT FALSE = 1.

Now we have the year, month, day of the month, hour, minute, and second calculated with adjustments for leap years.