splitRegex treats the regex as a delimiter, and does it fact "swallow" the match. Use matchRegexAll instead - e.g.:

1 import Text.Regex

  2 import Control.Applicative

  3

  4 main :: IO()

  5 main = do

  6     myIn <- readFile "Data.dat"

  7     case (intoEachPt myIn) of

  8         Nothing -> print "No match"

  9         Just (a, b, c, d) -> do print a

 10                                 print b

 11                                 print c

 12

 13 intoEachPt :: String -> Maybe (String, String, String, [String])

 14 intoEachPt = matchRegexAll (mkRegex "20[0-9]{13}AH021")

On Sun, Jul 28, 2013 at 5:35 AM, S. H. Aegis <shaegis@gmail.com> wrote:

Hi.
I'm newbie to Haskell.
I want to get date, it's type is [[String]].
The problem is that "the string that is used in regular expression pattern" is consumed. ie, disappear.
Here is my code.

------------------------------------------------------------
import Text.Regex
import Control.Applicative

main :: IO()
main = do
    myIn <- readFile "Data.dat"
    print $ lines <$> intoEachPt myIn

intoEachPt :: String -> [String]
intoEachPt = splitRegex (mkRegex "20[0-9]{13}AH021")
-----------------------------------------------------------

How can I fix this?


Data:
....there is many DIGIT.....201306000300001AH02112361640             9.......

Output:
[[....there is many DIGIT..."],["12361640             9......]....]

I hope:
[[....there is many DIGIT..."],["201306000300001AH02112361640             9......]....]

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