Question about Regular expression pattern consuming.
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......]....]
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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Wow! It's cool. One more question, please... In the DATA file, there is several 20[0-9]{13}AH201 strings. How can I treat all of each pattern? Thank you. Sincirely, S. Chang. 2013/7/29 Julian Arni <julian@edx.org>
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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participants (2)
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Julian Arni -
S. H. Aegis