[mick88]

Hi Jochemd, I think it's great that you decided to make this useful include.

I had a quick look at your code and I think it doesn't support negative timestamp. I made something like this the other day in C as a part of my programming assignment and I wanted to port it to PAWN, but seeing as you are already working on it, I may as well share my C code here. It supports negative timestamps, but doesn't regard timezones, but it's easy to implement.

Код:

bool is_leap_year(int year)
{
	/*
		This function checks if given year is a leap year: 
		it must be divisible by 4 and not divisible by 100 unless it's also divisible by 400
		- according to wikipedia
	*/
	return ((year % 4) == 0 && (((year % 100) != 0) || ((year % 400) == 0)));
}

int get_num_days(int year)
{
	/*
		Simple function, just returns number of days in given year: either 365 or 366 (leap year)
	*/
	return (is_leap_year(year)?366:365);
}

void timestamp_to_date(int timestamp, int *y, int *m, int *d, int *h, int *min, int *s)
{
	/*
		This is a highly sophisticated function that I created. It turns timestamp into ymd and hms.
		
	*/
	int days = timestamp / (60*60*24), //converting timestamp to days by dividing it by number of seconds in a day
		month_days[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; //defining number of days in a month
	/*
		Initialising starting date
	*/	

	if (timestamp >= 0)
	{
		/*
			Getting time of the day is very simple, So I do it first.
			Just take the ramainder from dividing by number of seconds in a day
		*/
		*s = timestamp % (60*60*24);

		*min = *s / 60;		//convert seconds to minutes...
		*s = *s % 60;		//...and keep the remainder

		*h = *min / 60;		//Same goes for hours...
		*min = *min % 60;	//...and minutes
	
		/*
			Now the ugly part: 'day' stores number of days since 1st January 1970.
			Somehow this has to be converted into a date.
			It's tricky, since some years have 366 days, some have 365.
		*/

		*d = 1;
		*m = 1;
		*y = 1970;

		int year_days;	//this variable stores number of days in current year, so I don't have to call the function twice in each iteration (efficiency)
		while (days > (year_days = get_num_days(*y)))
		{
			/*
				And loop through all years, substracting number of days passed each year...
				.. until number of days left is less than number of days in that year
			*/
			days -= year_days;
			 (*y)++;
		}

		/*
			Now tha I know what year it is, time to ensure that number of days in february is correct for the current year:
		*/
		if (is_leap_year(*y)) month_days[1] = 29;

		while (days > month_days[(*m)-1])
		{
			/*
				Looking for the month the same way I was looking for year
			*/
			days -=  month_days[(*m)-1];
			(*m)++;
		}
		// Remaining number of days is obviously the day of the month.
		*d += days;
	}
	else //same thing but in reverse
	{
		int year_days;
		days = -days;

		*s = -((timestamp) % (60*60*24))-1;

		*min = *s / 60;
		*s = *s % 60;

		*h = *min / 60;
		*min = *min % 60;
		
		*s = 59 - (*s);
		*min = 59-(*min);
		*h = 23-(*h);
		
		*d = 31;
		*m = 12;
		*y = 1969;

		while (days > (year_days = get_num_days(*y)))
		{
			days -= year_days;
			(*y)--;
		}

		if (is_leap_year(*y)) month_days[1] = 29;

		while (days > month_days[(*m)-1])
		{
			days -=  month_days[(*m)-1];
			(*m)--;
		}
		*d =  month_days[(*m)-1] - days;
	}
}