如何将此正则表达式转换为 Megaparsec 解析器而不会弄得一团糟?

Question

考虑这个正则表达式：

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

如果我在foo/bar/baz=one,two上运行它，它会匹配并且子组捕获one.如果我在foo/bar/baz/bar/baz=three,four,five上运行它，它会匹配并且子组捕获three,four。

我知道如何将其转换为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","")]]

这两种方法都按照我想要的方式工作。但是当我试图将其直接音译为Megaparsec时，情况很糟糕：

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 默认情况下没有回溯。我试图通过将try粘在一堆不同的地方来解决这个问题，但我无法让它工作。最终，我让这个怪物和notFollowedBy一起工作：

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"]

但这看起来一团糟！特别是，我不喜欢我实际上不得不两次指定大部分模式。从技术上讲，这不等同于正则表达式^foo/(?:[^=/]|/(?!baz=))+/baz=((?:.(?![^,]*$))*),[^,]*$，而不是我最初的正则表达式吗？必须有一种更好的方法来编写解析器。我该怎么做？

---

编辑：我也尝试过这种方式，这也有效(不，它错误地接受了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，它似乎同样混乱，manyTill意味着它不再真正映射到任何正则表达式。

Answer 1

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，请考虑使用

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

这将解析逗号分隔的内容列表。然后在纯代码中在该列表的最后一个元素之间重新插入逗号(例如，使用放置良好的fmap(。

从评论中可以看出，您似乎也遇到了这部分问题：

/[^=]+/baz=

你可以考虑这样的东西作为翻译：

slashPath = string "/" <++> path
path = string "baz=" <|> (many (noneOf "=/") <++> slashPath)
(<++>) = liftA2 (++)