How to convert epoch to gregorian datetime in haskell

Question

Let's say I have an integer representing Epoch time, for example epoch = 1499055085, and I want to convert it to UTCTime in Haskell. How can I do this?

In other languages this is a very trivial task, why is it so difficult in haskell?

Answer 1

Who says it is difficult?

You can simply use fromIntegral :: (Integral a, Num b) => a -> b to convert the epoch (an integer) to POSIXTime.

Next we can use posixSecondsToUTCTime :: POSIXTime -> UTCTime to obtain the UTCTime, and finally we use the utctDay :: UTCTime -> Day to get the day part of the UTCTime.

In case you want a (year,month,day) tuple, we can use the toGregorian :: Day -> (Integer,Int,Int) method.

So we can use the following method:

import Data.Time.Calendar(toGregorian)
import Data.Time.Clock(utctDay,UTCTime)
import Data.Time.Clock.POSIX(posixSecondsToUTCTime)

epochToUTC :: Integral a => a -> UTCTime
epochToUTC = posixSecondsToUTCTime . fromIntegral

epochToGregorian :: Integral a => a -> (Integer,Int,Int)
epochToGregorian = toGregorian . utctDay . epochToUTC

And then for instance:

Main> epochToGregorian 1234567
(1970,1,15)
Main> epochToGregorian 123456789
(1973,11,29)
Main> epochToGregorian 1234567890
(2009,2,13)