[Haskell-beginners] Question about Regular expression pattern consuming.

[Julian Arni]

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 at 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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>
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