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

[S. H. Aegis]

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 at 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 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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-- 
Sok Ha, CHANG
Dr. Chang's Clinic. #203. 503-23. AmSa-Dong, GangDong-Gu, Seoul.
Tel: +82-2-442-7585
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