How to convert seconds from an era to a current date and time?

Question

I know that I posted this some time ago, but I understood the solution. I wrote this code for the Roblox game, but I just post the code here if someone who has the same problem needs a solution. Anyway, here is the code:

outputTime = true -- true: will print the current time to output window. false: won't print time createVariable = true -- true: creates variables under game.Lighting. false: won't create variables ----------------------------------------------------------------------------------------------- --DO NOT EDIT BELOW---------------------------------------------------------------------------- ----------------------------------------------------------------------------------------------- if(createVariable) then yearVar = Instance.new("IntValue", game.Lighting) yearVar.Name = "Year" yearVar.Value = 0 monthVar = Instance.new("IntValue", game.Lighting) monthVar.Name = "Month" monthVar.Value = 0 dayVar = Instance.new("IntValue", game.Lighting) dayVar.Name = "Day" dayVar.Value = 0 hourVar = Instance.new("IntValue", game.Lighting) hourVar.Name = "Hour" hourVar.Value = 0 minuteVar = Instance.new("IntValue", game.Lighting) minuteVar.Name = "Minute" minuteVar.Value = 0 secondVar = Instance.new("IntValue", game.Lighting) secondVar.Name = "Second" secondVar.Value = 0 dayOfWeek = Instance.new("StringValue", game.Lighting) dayOfWeek.Name = "DayOfWeek" dayOfWeek.Value = "Thursday" end function giveZero(data) if string.len(data) <= 1 then return "0" .. data else return data end end function hasDecimal(value) if not(value == math.floor(value)) then return true else return false end end function isLeapYear(year) if(not hasDecimal(year / 4)) then if(hasDecimal(year / 100)) then return true else if(not hasDecimal(year / 400)) then return true else return false end end else return false end end local eYear = 1970 local timeStampDayOfWeak = 5 local secondsInHour = 3600 local secondsInDay = 86400 local secondsInYear = 31536000 local secondsInLeapYear = 31622400 local monthWith28 = 2419200 local monthWith29 = 2505600 local monthWith30 = 2592000 local monthWith31 = 2678400 local monthsWith30 = {4, 6, 9, 11} local monthsWith31 = {1, 3, 5, 7, 8, 10, 12} local daysSinceEpoch = 0 local DOWAssociates = {"Tursday", "Friday", "Saturday", "Sunday", "Monday", "Tuesday", "Wednesday"} while(true) do now = tick() year = 1970 secs = 0 daysSinceEpoch = 0 while((secs + secondsInLeapYear) < now or (secs + secondsInYear) < now) do if(isLeapYear(year+1)) then if((secs + secondsInLeapYear) < now) then secs = secs + secondsInLeapYear year = year + 1 daysSinceEpoch = daysSinceEpoch + 366 end else if((secs + secondsInYear) < now) then secs = secs + secondsInYear year = year + 1 daysSinceEpoch = daysSinceEpoch + 365 end end end secondsRemaining = now - secs monthSecs = 0 yearIsLeapYear = isLeapYear(year) month = 1 -- January while((monthSecs + monthWith28) < secondsRemaining or (monthSecs + monthWith30) < secondsRemaining or (monthSecs + monthWith31) < secondsRemaining) do if(month == 1) then if((monthSecs + monthWith31) < secondsRemaining) then month = 2 monthSecs = monthSecs + monthWith31 daysSinceEpoch = daysSinceEpoch + 31 else break end end if(month == 2) then if(not yearIsLeapYear) then if((monthSecs + monthWith28) < secondsRemaining) then month = 3 monthSecs = monthSecs + monthWith28 daysSinceEpoch = daysSinceEpoch + 28 else break end else if((monthSecs + monthWith29) < secondsRemaining) then month = 3 monthSecs = monthSecs + monthWith29 daysSinceEpoch = daysSinceEpoch + 29 else break end end end if(month == 3) then if((monthSecs + monthWith31) < secondsRemaining) then month = 4 monthSecs = monthSecs + monthWith31 daysSinceEpoch = daysSinceEpoch + 31 else break end end if(month == 4) then if((monthSecs + monthWith30) < secondsRemaining) then month = 5 monthSecs = monthSecs + monthWith30 daysSinceEpoch = daysSinceEpoch + 30 else break end end if(month == 5) then if((monthSecs + monthWith31) < secondsRemaining) then month = 6 monthSecs = monthSecs + monthWith31 daysSinceEpoch = daysSinceEpoch + 31 else break end end if(month == 6) then if((monthSecs + monthWith30) < secondsRemaining) then month = 7 monthSecs = monthSecs + monthWith30 daysSinceEpoch = daysSinceEpoch + 30 else break end end if(month == 7) then if((monthSecs + monthWith31) < secondsRemaining) then month = 8 monthSecs = monthSecs + monthWith31 daysSinceEpoch = daysSinceEpoch + 31 else break end end if(month == 8) then if((monthSecs + monthWith31) < secondsRemaining) then month = 9 monthSecs = monthSecs + monthWith31 daysSinceEpoch = daysSinceEpoch + 31 else break end end if(month == 9) then if((monthSecs + monthWith30) < secondsRemaining) then month = 10 monthSecs = monthSecs + monthWith30 daysSinceEpoch = daysSinceEpoch + 30 else break end end if(month == 10) then if((monthSecs + monthWith31) < secondsRemaining) then month = 11 monthSecs = monthSecs + monthWith31 daysSinceEpoch = daysSinceEpoch + 31 else break end end if(month == 11) then if((monthSecs + monthWith30) < secondsRemaining) then month = 12 monthSecs = monthSecs + monthWith30 daysSinceEpoch = daysSinceEpoch + 30 else break end end end day = 1 -- 1st daySecs = 0 daySecsRemaining = secondsRemaining - monthSecs while((daySecs + secondsInDay) < daySecsRemaining) do day = day + 1 daySecs = daySecs + secondsInDay daysSinceEpoch = daysSinceEpoch + 1 end hour = 0 -- Midnight hourSecs = 0 hourSecsRemaining = daySecsRemaining - daySecs while((hourSecs + secondsInHour) < hourSecsRemaining) do hour = hour + 1 hourSecs = hourSecs + secondsInHour end minute = 0 -- Midnight minuteSecs = 0 minuteSecsRemaining = hourSecsRemaining - hourSecs while((minuteSecs + 60) < minuteSecsRemaining) do minute = minute + 1 minuteSecs = minuteSecs + 60 end second = math.floor(now % 60) year = giveZero(year) month = giveZero(month) day = giveZero(day) hour = giveZero(hour) minute = giveZero(minute) second = giveZero(second) remanderForDOW = daysSinceEpoch % 7 DOW = DOWAssociates[remanderForDOW + 1] if(createVariable) then yearVar.Value = year monthVar.Value = month dayVar.Value = day hourVar.Value = hour minuteVar.Value = minute secondVar.Value = second dayOfWeek.Value = DOW end if(outputTime) then str = "Year: " .. year .. ", Month: " .. month .. ", Day: " .. day .. ", Hour: " .. hour .. ", Minute: " .. minute .. ", Second: ".. second .. ", Day of Week: " .. DOW print(str) end wait(1) end 

---- ORIGINAL MAIL ---- What are the formulas for calculating the following data without resources, except seconds from the moment of Epoch?

Here is a list of what I need:

Current month of the year Ex: 7

Current day of the month Ex: 25

Current day of the week Ex: Thursday (1-7 would be acceptable)

Current hour of the day Ex: 22

Current minute of an hour Example: 34

Current second of minute: 07

Answer 1

Here is what Lua code is adapted from some C code found by Google. It does not handle time slots or daylight saving time, so the outputs relate to universal coordinated time (UTC).

 -- based on http://www.ethernut.de/api/gmtime_8c_source.html local floor=math.floor local DSEC=24*60*60 -- secs in a day local YSEC=365*DSEC -- secs in a year local LSEC=YSEC+DSEC -- secs in a leap year local FSEC=4*YSEC+DSEC -- secs in a 4-year interval local BASE_DOW=4 -- 1970-01-01 was a Thursday local BASE_YEAR=1970 -- 1970 is the base year local _days={ -1, 30, 58, 89, 119, 150, 180, 211, 242, 272, 303, 333, 364 } local _lpdays={} for i=1,2 do _lpdays[i]=_days[i] end for i=3,13 do _lpdays[i]=_days[i]+1 end function gmtime(t) print(os.date("!\n%c\t%j",t),t) local y,j,m,d,w,h,n,s local mdays=_days s=t -- First calculate the number of four-year-interval, so calculation -- of leap year will be simple. Btw, because 2000 IS a leap year and -- 2100 is out of range, this formula is so simple. y=floor(s/FSEC) s=sy*FSEC y=y*4+BASE_YEAR -- 1970, 1974, 1978, ... if s>=YSEC then y=y+1 -- 1971, 1975, 1979,... s=s-YSEC if s>=YSEC then y=y+1 -- 1972, 1976, 1980,... (leap years!) s=s-YSEC if s>=LSEC then y=y+1 -- 1971, 1975, 1979,... s=s-LSEC else -- leap year mdays=_lpdays end end end j=floor(s/DSEC) s=sj*DSEC local m=1 while mdays[m]<j do m=m+1 end m=m-1 local d=j-mdays[m] -- Calculate day of week. Sunday is 0 w=(floor(t/DSEC)+BASE_DOW)%7 -- Calculate the time of day from the remaining seconds h=floor(s/3600) s=sh*3600 n=floor(s/60) s=sn*60 print("y","j","m","d","w","h","n","s") print(y,j+1,m,d,w,h,n,s) end local t=os.time() gmtime(t) t=os.time{year=1970, month=1, day=1, hour=0} gmtime(t) t=os.time{year=1970, month=1, day=3, hour=0} gmtime(t) t=os.time{year=1970, month=1, day=2, hour=23-3, min=59, sec=59} gmtime(t) 

Answer 2

The formula is not simple for several reasons, especially leap years. You should probably use the date function on this page , rather than trying to figure it out yourself.

Answer 3

You can use luatz

 x = 1234567890 t = require "luatz.timetable".new_from_timestamp ( x ) print(t.year,t.month,t.day,t.hour,t.min,t.sec,t.yday,t.wday) -- Or just print(t:rfc_3339()) 

Answer 4

It would be much faster to use my method, which I actually did not see, as someone else is using access to os.date ()

Since I do not have access to os.date (), here is my solution:

 local tabIndexOverflow = function(seed, table) -- This subtracts values from the table from seed until an overflow -- This can be used for probability :D for i = 1, #table do if seed - table[i] <= 0 then return i, seed end seed = seed - table[i] end end local getDate = function(unix) -- Given unix date, return string date assert(unix == nil or type(unix) == "number" or unix:find("/Date%((%d+)"), "Please input a valid number to \"getDate\"") local unix = (type(unix) == "string" and unix:match("/Date%((%d+)") / 1000 or unix or os.time()) -- This is for a certain JSON compatability. It works the same even if you don't need it local dayCount, year, days, month = function(yr) return (yr % 4 == 0 and (yr % 100 ~= 0 or yr % 400 == 0)) and 366 or 365 end, 1970, math.ceil(unix/86400) while days >= dayCount(year) do days = days - dayCount(year) year = year + 1 end -- Calculate year and days into that year month, days = tabIndexOverflow(days, {31,(dayCount(year) == 366 and 29 or 28),31,30,31,30,31,31,30,31,30,31}) -- Subtract from days to find current month and leftover days -- hours = hours > 12 and hours - 12 or hours == 0 and 12 or hours -- Change to proper am or pm time -- local period = hours > 12 and "pm" or "am" -- Formats for you! -- string.format("%d/%d/%04d", month, days, year) -- string.format("%02d:%02d:%02d %s", hours, minutes, seconds, period) return {Month = month, day = days, year = year, hours = math.floor(unix / 3600 % 24), minutes = math.floor(unix / 60 % 60), seconds = math.floor(unix % 60)} end 

You, however, are on their own when it comes to finding the day of the week. I never cared whether it would be Thor day or Frig’s day.

Answer 5

os.date is a standard Lua function, if you pass the first argument as "%t" , it will return a table containing the following fields: year (four digits), month (1-12), day (1-31), hour ( 0-23), min (0--59), sec (0--61), wday (weekday, Sunday - 1), yday (day of the year) and isdst (daylight saving time flag, logical).

Give him a test:

 time = os.time() print("time since epoch: " .. time) date = os.date("*t", time) print("year: " .. date.year) print("month: " .. date.month) print("day: " .. date.day) print("hour: " .. date.hour) print("minute: " .. date.min) print("second: " .. date.sec) print("weekday: " .. date.wday) 

Output:

 time since epoch: 1374826427 year: 2013 month: 7 day: 26 hour: 16 minute: 13 second: 47 weekday: 6 

Answer 6

This is how I do it.

 > time0=os.time() > time0 1571439964 > os.date("%Y%m%d%H%M%S",time0) 20191019120604 >