在parsec中的单引号字符串中解析单个qoute字符 [英] Parsing single qoute char in a single-quoted string in parsec

问题描述

我的parsec解析器中有一个愚蠢的情况，我希望得到您的帮助。

我需要解析一系列由|分隔的强/字符。字符。
因此，我们可以有一个| b |'c'|'abcd'

这应该变成

  [a，b，c，abcd] 
  

空格不允许，除非在''字符串中。现在，在我天真的尝试中，我现在已经得到了可以将字符串解析为[a'a，bb]而不是aa |'b'b'的字符串[aa，b'b] 。

  singleQuotedChar :: Parser Char 
 singleQuotedChar = noneOf'< |> try（string''>> return'\''）
 
 simpleLabel = do 
 whiteSpace haskelldef 
 lab<  -  many1（noneOf|） 
 return $ lab 
 
 quotedLabel = do 
 whiteSpace haskelldef 
 char'\''
 lab<  -  many singleQuotedChar 
 char' \''
 return $ lab 
  

现在，我该如何告诉解析器考虑如果它后面跟着一个|，那么'停止'或空白？
（或者，得到一些'字符计入此）。输入是用户生成的，所以我不能依赖他们\\-chars。

由引号分隔的字符串的中间部分非常混乱，但我相信这应该允许您解析它。

  quotedLabel = do  - 读取第一个报价。 
 whiteSpace 
 char'\''
 quotedLabel2 
 
 quotedLabel2 = do  - 读取字符串和结束报价。 
 lab < - 许多singleQuotedChar 
尝试（做更多<  -  quotedLabel3 
 return $ lttracequotedLabel2（lab ++ more））
< |> （do char'\''
 return $ lttracequotedLabel2lab）
 
 
 quotedLabel3 = do  - 处理中间引号
 char'\'' 
 lookAhead $ noneOf ['|'] 
 ret<  -  quotedLabel2 
 return $ lttracequotedLabel3$'++ ret 
  

I've got a silly situation in my parsec parsers that I would like your help on.

I need to parse a sequence of strongs / chars that are separated by | characters. So, we could have a|b|'c'|'abcd'

which should be turned into

[a,b,c,abcd]

Space is not allowed, unless inside of a ' ' string. Now, in my naïve attempt, I got the situation now where I can parse strings like a'a|'bb' to [a'a,bb] but not aa|'b'b' to [aa,b'b].

singleQuotedChar :: Parser Char
singleQuotedChar = noneOf "'" <|> try (string "''" >> return '\'')

simpleLabel = do
    whiteSpace haskelldef
    lab <- many1 (noneOf "|")
    return $ lab

quotedLabel = do
    whiteSpace haskelldef
    char '\''
    lab <- many singleQuotedChar
    char '\''
    return $ lab

Now, how do I tell the parser to consider ' a stoping ' iff it is followed by a | or white space? (Or, get some ' char counting into this). The input is user generated, so I cannot rely on them \'-ing chars.

解决方案

Note that allowing a quote in the middle of a string delimited by quotes is very confusing to read, but I believe this should allow you to parse it.

quotedLabel = do -- reads the first quote.
    whiteSpace
    char '\''
    quotedLabel2

quotedLabel2 = do -- reads the string and the finishing quote.
    lab <- many singleQuotedChar
    try  (do more <- quotedLabel3
             return $ lttrace "quotedLabel2" (lab ++ more))
     <|> (do char '\''
             return $ lttrace "quotedLabel2" lab)


quotedLabel3 = do -- handle middle quotes
    char '\''
    lookAhead $ noneOf ['|']
    ret <- quotedLabel2
    return $ lttrace "quotedLabel3" $ "'" ++ ret

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