Question

我认为我已经设法将字符串解析为字符串和字符串到Ints但我还需要解析（String，Int）类型，因为userRatings是为了正确读取textFile而我正在使用Parsec

这是解析和导入

import Text.Parsec
( Parsec, ParseError, parse        -- Types and parser
, between, noneOf, sepBy, many1    -- Combinators
, char, spaces, digit, newline     -- Simple parsers
)

-- Parse a string to a string
stringLit :: Parsec String u String
stringLit = between (char '"') (char '"') $ many1 $ noneOf "\"\n"

-- Parse a string to a list of strings
listOfStrings :: Parsec String u [String]
listOfStrings = stringLit `sepBy` (char ',' >> spaces)

-- Parse a string to an int
intLit :: Parsec String u Int
intLit = fmap read $ many1 digit
-- Or `read <$> many1 digit` with Control.Applicative

film :: Parsec String u Film
film = do
-- alternatively `title <- stringLit <* newline` with Control.Applicative
title <- stringLit
newline
director <- stringLit
newline
year <- intLit
newline
userRatings <- listOfStrings
newline
return (title, director, year, userRatings)

Answer 1

您可以通过使用Applicative或Monad接口从现有解析器进行合成来完成此操作。 Parser包含两者的实例。

使用Applicative：

stringIntTuple :: Parser (String, Int)
stringIntTuple = (,) <$> yourStringParser <*> yourIntParser

与以下内容相同：

stringIntTuple :: Parser (String, Int)
stringIntTuple = liftA2 (,) yourStringParser yourIntParser

使用Monad：

stringIntTuple :: Parser (String, Int)
stringIntTuple =
  do
    theString <- yourStringParser
    theInt <- yourIntParser
    return (theString, theInt)

如何使用parsec解析Haskell中的元组（String，Int）

1 个答案: