Thanks. import Text.ParserCombinators.Parsec simple = between (char '"') (char '"') (many (satisfy (/= '"'))) multiple = sepBy simple (char ',') total = sepBy multiple (char '@') str2lsts :: String -> [[String]] str2lsts str= case parse total "" str of Left err -> error (show err) Right lsts -> lsts solves my problem. As you can see I've replaced '§' with '@' and now all works fine. Luca.

From: daniel.is.fischer@web.de To: beginners@haskell.org Subject: Re: [Haskell-beginners] beginner question Date: Fri, 30 Oct 2009 15:46:33 +0100

Am Freitag 30 Oktober 2009 14:40:13 schrieb Luca Ciciriello:

...

Hi all.

Just a very basic question.

I need to write a function str2lsts :: String -> [[String]] in order to transorm a string like:

"\"1\",\"cat\",\"dog\"§\"2\",\"duck\",\"goose\""

in the list of lists:

[["1","cat","dog"],["2","duck","goose"]]

I've tried to mix recursion, pattern matching and list comprehension, but the obtained result was embarrassing complex (> 20 lines of awful code). I think that a more simple solution certainly exists.

splitOnToken :: Eq a => a -> [a] -> [[a]] splitOnToken t xs = case break (== t) xs of (hd,tl) -> hd:case tl of (_:r@(_:_)) -> splitOnToken t r _ -> []

str2lsts = map (map read . splitOnToken ',') . splitOnToken '§'

if things weren't enclosed in quotation marks inside the string, it would be the nicer

map (splitOnToken ',') . splitOnToken '§'

, provided of course, neither ',' nor '§' are valid characters for the target strings.

import Text.ParserCombinators.Parsec

simple = between (char '"') (char '"') (many (staisfy (/= '"'))) -- alternative: simple = char '"' >> manyTill anyChar (char '"')

multiple = sepBy simple (char ',')

total = sepBy multiple (char '§')

str2lsts str = case parse total "" str of Left err -> error (show err) Right lsts -> lsts

...

Thanks in advance for any idea.

Luca

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