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考虑这个正则表达式：

^foo/[^=]+/baz=(.*),[^,]*$

如果我运行它foo/bar/baz=one,two，它匹配并且子组捕获one。如果我运行它foo/bar/baz/bar/baz=three,four,five，它匹配并且子组捕获three,four。

我知道如何把它变成regex-applicative解析器或ReadP解析器：

import Text.Regex.Applicative
match (string "foo/" *> some (psym (/= '=')) *> string "/baz=" *> many anySym <* sym ',' <* many (psym (/= ','))) <$> ["foo/bar/baz=one,two", "foo/bar/baz/bar/baz=three,four,five"]
-- [Just "one",Just "three,four"]

import Text.ParserCombinators.ReadP
readP_to_S (string "foo/" *> many1 (satisfy (/= '=')) *> string "/baz=" *> many get <* char ',' <* many (satisfy (/= ',')) <* eof) <$> ["foo/bar/baz=one,two", "foo/bar/baz/bar/baz=three,four,five"]
-- [[("one","")],[("three,four","")]]

这两个都按照我想要的方式工作。但是当我尝试将其直接音译成 Megaparsec 时，它变得很糟糕：

import Text.Megaparsec
parse (chunk "foo/" *> some (anySingleBut '=') *> chunk "/baz=" *> many anySingle <* single ',' <* many (anySingleBut ',') <* eof) "" <$> ["foo/bar/baz=one,two", "foo/bar/baz/bar/baz=three,four,five"]
-- [Left (ParseErrorBundle {bundleErrors = TrivialError 11 (Just (Tokens ('=' :| "one,"))) (fromList [Tokens ('/' :| "baz=")]) :| [], bundlePosState = PosState {pstateInput = "foo/bar/baz=one,two", pstateOffset = 0, pstateSourcePos = SourcePos {sourceName = "", sourceLine = Pos 1, sourceColumn = Pos 1}, pstateTabWidth = Pos 8, pstateLinePrefix = ""}}),Left (ParseErrorBundle {bundleErrors = TrivialError 19 (Just (Tokens ('=' :| "thre"))) (fromList [Tokens ('/' :| "baz=")]) :| [], bundlePosState = PosState {pstateInput = "foo/bar/baz/bar/baz=three,four,five", pstateOffset = 0, pstateSourcePos = SourcePos {sourceName = "", sourceLine = Pos 1, sourceColumn = Pos 1}, pstateTabWidth = Pos 8, pstateLinePrefix = ""}})]

我知道这源于 Megaparsec 默认情况下不回溯。我试图通过坚持try在一堆不同的地方来解决这个问题，但我无法让它发挥作用。最终，我得到了这个怪物notFollowedBy：

import Text.Megaparsec
parse (chunk "foo/" *> some (noneOf "=/" <|> try (single '/' <* notFollowedBy (chunk "baz="))) *> chunk "/baz=" *> many (try (anySingle <* notFollowedBy (many (anySingleBut ',') <* eof))) <* single ',' <* many (anySingleBut ',') <* eof) "" <$> ["foo/bar/baz=one,two", "foo/bar/baz/bar/baz=three,four,five"]
-- [Right "one",Right "three,four"]

但这看起来像一团糟！特别是，我不喜欢我实际上必须两次指定大部分模式。从技术上讲，这不等于 regex ^foo/(?:[^=/]|/(?!baz=))+/baz=((?:.(?![^,]*$))*),[^,]*$，而不是我最初的 regex 吗？必须有更好的方法来编写该解析器。我该怎么做？

编辑：我也尝试过这种方式，它也有效（不，它错误地接受foo//baz=,）：

import Text.Megaparsec
parse (chunk "foo/" *> (some . try $ many (noneOf "=/") *> single '/') *> chunk "baz=" *> ((++) <$> many (anySingleBut ',') <*> (concat <$> manyTill ((:) <$> single ',' <*> many (anySingleBut ',')) (try $ single ',' *> many (anySingleBut ',') *> eof)))) "" <$> ["foo/bar/baz=one,two", "foo/bar/baz/bar/baz=three,four,five"]
-- [Right "one",Right "three,four"]

不过，它看起来一样混乱，这manyTill意味着它不再真正映射到任何正则表达式。

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1 回答 1

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如果不仔细阅读，我想给你带来麻烦的是这部分：

(.*),[^,]*

如果是这样，那么考虑使用

sepBy (many (noneOf ",")) (string ",")

它将解析逗号分隔的事物列表。然后在除最后一个元素之外的所有元素之间以纯代码重新插入逗号（例如，放置良好的fmap）。

从评论中，您似乎在这部分也遇到了一些问题：

/[^=]+/baz=

你可以考虑这样的翻译：

slashPath = string "/" <++> path
path = string "baz=" <|> (many (noneOf "=/") <++> slashPath)
(<++>) = liftA2 (++)
于 2020-01-08T21:39:13.067 回答