Question

我正在尝试解析RFC5322电子邮件地址。我的解析器的工作原理是，在结果中，其中一个是正确的。但是，我该如何选择“正确”的结果？

给定字符串Foo Bar <foo@bar.com>，我的解析器应该生成Address (Just "Foo Bar") "foo@bar.com"的值。

或者，给定字符串foo@bar.com，我的解析器应生成Address Nothing "foo@bar.com"的值。

首选包含名称的值。

我的解析器看起来像这样：

import           Control.Applicative
import           Data.Char
import qualified Data.Text                     as T
import           Text.ParserCombinators.ReadP

onlyEmail :: ReadP Address
onlyEmail = do
  skipSpaces
  email <- many1 $ satisfy isAscii
  skipSpaces
  return $ Address Nothing (T.pack email)

withName :: ReadP Address
withName = do
  skipSpaces
  name <- many1 (satisfy isAscii)
  skipSpaces
  email <- between (char '<') (char '>') (many1 $ satisfy isAscii)
  skipSpaces
  return $ Address (Just $ T.pack name) (T.pack email)

rfc5322 :: ReadP Address
rfc5322 = withName <|> onlyEmail

当我使用readP_to_S rfc5322 "Foo Bar <foo@bar.com>"运行解析器时，它会产生以下结果：

[ (Address {addressName = Nothing, addressEmail = "F"},"oo Bar <foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Fo"},"o Bar <foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo"},"Bar <foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo "},"Bar <foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo B"},"ar <foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Ba"},"r <foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar"},"<foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar "},"<foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <"},"foo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <f"},"oo@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <fo"},"o@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo"},"@bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@"},"bar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@b"},"ar.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@ba"},"r.com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@bar"},".com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@bar."},"com>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@bar.c"},"om>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@bar.co"},"m>")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@bar.com"},">")
, (Address {addressName = Just "Foo Bar", addressEmail = "foo@bar.com"},"")
, (Address {addressName = Just "Foo Bar ", addressEmail = "foo@bar.com"},"")
, (Address {addressName = Nothing, addressEmail = "Foo Bar <foo@bar.com>"},"")
]

在这种情况下，我实际想要的结果在列表中显示为倒数第三。我该如何表达这种偏好？

Answer 1

你不应该偏好。您的问题是您的部分解析器正在接受比实际需要的更大的字符串集。

例如，我的解决方案：

import           Control.Bool
import           Control.Applicative
import           Data.Char
import qualified Data.Text                     as T
import           Data.Text (Text)
import           Text.ParserCombinators.ReadP

email :: ReadP Text
email = do
    l <- part
    a <- char '@'
    d <- part
    return . T.pack $ l ++ a:d
  where
    part = munch1 (isAscii <&&> (/='@') <&&> (/='<') <&&> (/='>'))

name :: ReadP Text
name = T.pack <$> chainr1 part sep
  where
    part = munch1 (isAlpha <||> isDigit <||> (=='\''))
    sep  = (\xs ys -> xs ++ ' ':ys) <$ munch1 (==' ')

onlyEmail :: ReadP Address
onlyEmail = Address Nothing <$> email

withName :: ReadP Address
withName = do
    n <- name
    skipSpaces
    e <- between (char '<') (char '>') email
    return $ Address (Just n) e

address :: ReadP Address
address = skipSpaces *> (withName <|> onlyEmail)

main = print $ readP_to_S address "Foo Bar <foo@bar.com>"

将被打印：

[(Address (Just "Foo Bar") "foo@bar.com","")]

选择正确的ReadP解析结果

1 个答案: