Module:Calendar

Module documentation[ create · purge ]

This documentation is transcluded from Module:Calendar/doc.

local p = {}

-- Convert "Julian day number" (jdn) to a calendar date
-- "gregorian" is a 1 for gregorian calendar and 0 for Julian
-- based on https://en.wikipedia.org/wiki/Julian_day#Converting_Julian_or_Gregorian_calendar_date_to_Julian_day_number
function p.jdn2date(jdn, gregorian)
  local f, e, g, h, year, month, day

	f = jdn + 1401
	if gregorian>0 then
		 f = f + math.floor((math.floor((4*jdn + 274277) / 146097) * 3) / 4) - 38
	end
  e = 4*f + 3
  g = math.floor(math.fmod(e, 1461) / 4)
  h = 5*g + 2
  day   = math.floor((math.fmod(h,153)) / 5) + 1
  month = math.fmod (math.floor(h/153) + 2, 12) + 1
  year  = math.floor(e/1461) - y + math.floor((12 + 2 - month) / 12) 
	
	-- If year is less than 1, subtract one to convert from a zero based date system to the
	-- common era system in which the year -1 (1 B.C.E) is followed by year 1 (1 C.E.).
	if year < 1 then
			year = year - 1
	end
	
	return year, month, day
end

-- Convert calendar date to  "Julian day number" (jdn) 
-- "gregorian" is a 1 for gregorian calendar and 0 for Julian
-- based on  https://en.wikipedia.org/wiki/Julian_day#Converting_Julian_or_Gregorian_calendar_date_to_Julian_day_number
-- explanation based on http://www.cs.utsa.edu/~cs1063/projects/Spring2011/Project1/project1.html
function p.date2jdn(year, month, day, gregorian)
	a = math.floor((14-month) / 12) -- will be 1 for January and February, and 0 for other months.
	y = year + 4800 - a             -- years since year –4800
	m = month + 12*a - 3            -- month number where 10 for January, 11 for February, 0 for March, 1 for April
	c = math.floor((153*m + 2)/5)   -- number of days since March 1
	d = 32045                       -- offset so the result will be 0 for January 1, 4713 BCE
	if gregorian>0 then	
		b = math.floor(y/4) - math.floor(y/100) + math.floor(y/400) -- number of leap years since y==0 (year –4800)
		d = 32045                     -- offset so the result will be 0 for January 1, 4713 BCE
	else
		b = math.floor(y/4)           -- number of leap years since y==0 (year –4800)
	  d = 32083                     -- offset so the result will be 0 for January 1, 4713 BCE
	end
	return day + c + 365*y + b - d
end

-- Convert a date from Gregorian to Julian calendar
function p.Gregorian2Julian(year, month, day)
	return jdn2date(date2jdn(year, month, day, 1), 0)
end

-- Convert a date from Julian to Gregorian calendar
function p.Julian2Gregorian(year, month, day)
	return jdn2date(date2jdn(year, month, day, 0), 1)
end

-- Return day of week based on gregorian date. Mon->1, Tue->2, ..., Sun->7
function p.DayOfWeek(year, month, day)
	return math.fmod(date2jdn(year, month, day, 1), 7) + 1
end

return p;