Parser in Haskell for Triples - parsing

Question

I am currently doing an assignment about Parsing in Haskell, but I am struggling with some of the basics.
Assignment :
I am supposed to create a function which parses a string into a list of Triples.
So that:
A, B, C
,E ,D
would result in
Triples [("A","B","C"), ("A","E","D")]
The input string is going to include ;\n as an indication for the beginng of a new Triple. The string is going to end with a dot.
The elements of the Triples can be letters or digits or combination,
e.g. abc, a, 1, abc121.
Therefore,
"a,b,c;\n d,e;\n f,g;\n h,i."
would result in:
Triples [("a","b","c"),("a","d","e"),("a","f","g"),("a","h","i")]
My current solution:
parseTriplesD :: Parser Triples
parseTriplesD = parseTriples
>>= \rs -> return (Triples rs)
This function is pretty simple and correct. Takes the string and returns a object of the newtype Triples with the List of Triples created by parseTriples.
parseTriples :: Parser [Triple]
parseTriples = parseTriple
>>= \r -> ((string ";\n" >> parseTriples >>= \rs -> return (r:rs))
P.<|>(return[r]))
This function needs some work. My idea is that I use another function which creates a Triple with tree Elements of the input string, ignores the /n and recursivly calls it self while adding the created triples to a return list. When this does not work because it can only create one Triple, it returns a list with the Triple.
I somehow need to create the first Triple, and then use first element of this triple as the first element of the other ones.
Question 1
How do I create the first Triple and use the first Elements of the Triple for the other Triples?
parseTriple :: Parser Triple
parseTriple = P.many (letter<|>digit) >>= \a -> P.char ','
>> P.many (letter<|>digit)>>= \b -> P.char ','
>> P.many (letter<|>digit)>>= \c -> return ((a,b,c))
This function is pretty simple but I am not sure if its correct.
My idea is that it takes the first couple of characters of the string which are either a letter or a digit, up until the comma "," , and saves these charcters in a.
It is repeated 3 times, and the creates and returns a Triple with the three elements.
Question 2
How do I take only a few characters (which are either a letter or a digit EDIT: Or A SPACE Character) of the string up until the comma?
Is P.many (letter<|>digit) correct?
What we are given:
The Triples data structue:
newtype Triples = Triples [Triple] deriving (Show,Eq)
type Triple = (String, String, String)
Imports:
import Test.HUnit (runTestTT,Test(TestLabel,TestList),(~?=))
import qualified Text.Parsec as P (char,runP,noneOf,many,(<|>),eof)
import Text.ParserCombinators.Parsec
import Text.Parsec.String
import Text.Parsec.Char
import Data.Maybe
Test cases
runParsec :: Parser a -> String -> Maybe a
runParsec parser input = case P.runP parser () "" input of
Left _ -> Nothing
Right a -> Just a
-- | Tests the implementations of 'parseScore'.
main :: IO ()
main = do
testresults <- runTestTT tests
print testresults
-- | List of tests for 'parseScore'.
tests :: Test
tests = TestLabel "parseScoreTest" (TestList [
runParsec parseTriplesD "0,1,2;\n2,3." ~?= Just (Triples [("0","1","2"),("0","2","3")]),
runParsec parseTriplesD "a,bcde ,23." ~?= Just (Triples [("a","bcde ","23")]),
runParsec parseTriplesD "a,b,c;\n d,e;\n f,g;\n h,i." ~?= Just (Triples [("a","b","c"),("a","d","e"),("a","f","g"),("a","h","i")]),
runParsec parseTriplesD "a,bcde23." ~?= Nothing,
runParsec parseTriplesD "a,b,c;d,e;f,g;h,i." ~?= Nothing,
runParsec parseTriplesD "a,b,c;\nd;\nf,g;\nh,i." ~?= Nothing
])

Answer 1

What you could do is:
Parse the first character
Parse a list of pairs
Add the first character to each of the pairs to create triples
Using do notation will make your code more readable.
You can use alphaNum as a shorthand for letter <|> digit.
parseTriplesD :: Parser Triples
parseTriplesD = Triples <$> parseTriples
parseTriples :: Parser [Triple]
parseTriples = do
a <- parseString
char ','
pairs <- parsePair `sepBy1` string ";\n"
char '.'
eof
return (map (\(b, c) -> (a, b, c)) pairs)
parsePair :: Parser (String, String)
parsePair = do
first <- parseString
char ','
second <- parseString
return (first, second)
parseString :: Parser String
parseString = many (char ' ') >> many (alphaNum <|> char ' ')