如何使用 parsec 在 Haskell 中解析元组（String，Int）

Question

我想我已经设法将字符串解析为字符串并将字符串解析为 Ints，但我还需要解析 (String,Int) 类型，因为 userRatings 是为了正确读取 textFile 并且我正在使用 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组合来做到这一点。 Parser 两者都有。

使用Applicative：

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

与以下相同：

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

使用Monad：

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