今日の日付を文字列で取得する。"2009/06/03"

Haskell には System.Time と Data.Time のふたつがあるようだ。

Prelude> :m + System.Time

Prelude System.Time> getClockTime

Wed Jun  3 09:45:04 東京 (標準時) 2009

getClockTime で取得した値を CalendarTime に。

 

Prelude System.Time> tm <- getClockTime

Prelude System.Time> toCalendarTime tm

CalendarTime {ctYear = 2009, ctMonth = June, ctDay = 3, ctHour = 9, ctMin = 45, 

ctSec = 48, ctPicosec = 456107000000, ct

WDay = Wednesday, ctYDay = 153, 

ctTZName = "\147\140\139\158 (\149W\143\128\142\158)", 

ctTZ = 32400, ctIsDST = False}

getClockTime で取得した値を UTCTime に。

 

Prelude System.Time> toUTCTime tm

CalendarTime {ctYear = 2009, ctMonth = June, ctDay = 3, ctHour = 0, ctMin = 45, 

ctSec = 48, ctPicosec = 456107000000, ctWDay = Wednesday, ctYDay = 153, 

ctTZName = "UTC", ctTZ = 0, ctIsDST = False}

こっちは Data.Time

Prelude> :m + Data.Time

Prelude Data.Time> :t getCurrentTime

getCurrentTime :: IO UTCTime

Prelude Data.Time> getCurrentTime

2009-06-03 00:47:48.476746 UTC

Data.Time の東京 (標準時)

Prelude Data.Time> :t getZonedTime

getZonedTime :: IO ZonedTime

 

Prelude Data.Time> getCurrentTimeZone

東京 (標準時)

 

Prelude Data.Time> getZonedTime

2009-06-03 09:50:15.2163775 東京 (標準時)

今日の日付を文字列で取得する。"2009/06/03"

 import Data.Time

 

 dateString :: IO [Char]

 dateString = do 

     zonedTm <- getZonedTime

     return $ replace '-' '/' $ take 10 $ show zonedTm

 

 -- replace from to xs

 replace :: (Eq a) => a -> a -> [a] -> [a]

 replace _ _ [ ] = [ ]

 replace from to (x:xs) | from == x = to : (replace  from to xs)

                        | otherwise = x  : (replace from to xs)

 *Japanese> :load Japanese.hs

 [1 of 1] Compiling Japanese         ( Japanese.hs, interpreted )

 Ok, modules loaded: Japanese.

 *Japanese> dateString

 "2009/06/03"