您好我试图编写一个将markdown转换为HTML的程序，我知道有Pandoc，但我的项目是手动编写的。我已经完成了或至少我认为我这样做但是我得到以下错误Haskell错误lexical error in string/character literal at character ' \r'我不知道指的是什么，指出它的任何帮助都会很棒。谢谢所以更新：我用misc符号更改了一些内容，我现在得到的错误是hs.38:17: Not in scope 'str'，当我正在处理它已反复指向第38行的事情时，我无法弄清楚问题是什么因为它忽略了前一个函数中的相同内容
module Main
(
convertToHTML,
convertSpecialChar,
main
) where
import System.Environment (getArgs)
import System.IO
import Data.Char
import Data.List
import Data.List.Split
eof = "</body></html>"
convertToHTML :: String -> String
convertToHTML x = specialTags $ headings $ endings $ beginnings $ replace "---"<hr>" x
convertSpecialChar :: String -> String
ConvertSpecialChar x = (convertLessThan $ convertAmpersand $ convertGreaterThan x)++eof
where
convertLessThan str = concat [if c =='<' then '<" else [c] | c <- str]
convertAmpersand str = concat [if c == '&' then "&" else [c] | c <- str]
convertGreaterThan str = concat [if c =='>' then ">" else [c] | c <- str]
beginnings :: String -> String
beginnings str = unwords $ map tag ch
where
tag x
| isPrefixOf "**" x = "<strong>" ++ (tail $ tail x)
| isPrefixOf "__" x = "<strong>" ++ (tail $ tail x)
| isPrefixOf "_" x = "<em>" ++ (tail x)
| isPrefixOf "*" x = "<em>" ++ (tail x)
| isPrefixOf "^" x = "<p>" ++ (tail x)
| isPrefixOf "---" x = replace "---" "<hr>"
| otherwise = x
ch =splitOn " " str
replace :: Eq a => [a] -> [a] -> [a] -> [a]
replace old new x = intercalate new (splitOn old x)
endings :: String -> String
endings str = unwords $ map tag ch
where
tag x
| isInfixOf "**" x = replace "**" "</strong>" x
| isInfixOf "__" x = replace "__" "</strong>" x
| isInfixOf "_" x = replace "_" "</em>" x
| isInfixOf "*" x = replace "*" "</em>" x
| isInfixOf "^" x = replace "^" "</p>" x
| isInfixOf "---" x = replace "---" "<hr>" x
| otherwise = x
ch = splitOn " " str
headings str = unlines $ map heads (lines str)
where
heads x
| isPrefixOf "######" x = "<h6>" ++ (numTail 6 x) ++ "</h6>"
| isPrefixOf "#####" x = "<h5>" ++ (numTail 5 x) ++ "</h5>"
| isPrefixOf "####" x = "<h4>" ++ (numTail 4 x) ++ "</h4>"
| isPrefixOf "###" x = "<h3>" ++ (numTail 3 x) ++ "</h3>"
| isPrefixOf "##" x = "<h2>" ++ (tail $ tail x) ++ "</h2>"
| isPrefixOf "#" x = "<h1>" ++ (tail x) ++ "</h1>"
| otherwise = x
specialTags str = unlines $ map tags (lines str)
where
tags x
| isPrefixOf "[code]" x = "<blockquote><pre><code>" ++ (numTail 6 x)
| isSuffixOf "[code]" x = (numInit 6 x) ++ "</code></pre></blockquote>"
| otherwise = x
numTail :: Int -> String -> String
numTail _ [] = []
numTail 1 str = tail str
numTail x str = tail $ (numTail (x-1) str)
numInit :: Int-> String -> String
numInit _ [] = []
numInit 1 str = init str
numInit x str = init $ (numInit (x-1) str)
main = do
args <- getArgs
let (infile,outfile) = (\\(x:y:ys) -> (x,y)) args
putStrLn $ "Input file: " ++ infile
putStrLn $ "Output file: " ++ outfile
contents <- readFile infile
let contentlines = unlines $ tail $ lines contents
let title = head $ lines contents
let header = "<!DOCTYPE html> <head>" ++ "<meta http-equiv = \\"Content-type\\"content=\\"text/html; charset=utf-8\\" />" ++ "<title>" ++title++"</title>" ++"</head><body>"
writeFile outfile $ convertToHTML $ header ++ convertSpecialChar contentlines
答案 0 :(得分：4)
在第17行，您缺少双引号。我想replace "---"<hr>" x应该阅读replace "---" "<hr>" x。
您的函数convertToHTML的整个声明将会读取
convertToHTML :: String -> String
convertToHTML x = specialTags $ headings $ endings $ beginnings $ replace "---" "<hr>" x
'\r'？编译Haskell模块的第一个阶段是词法分析，其中您的程序文本被分解为令牌，然后由解析器处理。在您的情况下，词法分析失败，因为假定新的字符串文字由" x启动。然后它遇到第17行的结尾，没有文字开头正确终止，结尾双引号或字符串文字的任何指示是multiline string literal。由于这是非法的词法语法，因此它抱怨遇到行尾（'\r'）。
不可否认，如果错误消息明确提到非终止字符串文字，那将会更有帮助。
在任何情况下，具有语法突出显示支持Haskell的编辑器可能早就暗示了这个问题。 ;）
您在变量str不在ch的本地定义范围内时遇到的问题是由于布局造成的。确保ch缩进到与前一个tag定义相同的级别。也就是说，而不是
beginnings :: String -> String
beginnings str = unwords $ map tag ch
where
tag x
| isPrefixOf "**" x = "<strong>" ++ (tail $ tail x)
| isPrefixOf "__" x = "<strong>" ++ (tail $ tail x)
| isPrefixOf "_" x = "<em>" ++ (tail x)
| isPrefixOf "*" x = "<em>" ++ (tail x)
| isPrefixOf "^" x = "<p>" ++ (tail x)
| isPrefixOf "---" x = replace "---" "<hr>"
| otherwise = x
ch =splitOn " " str
你应该写点像
beginnings :: String -> String
beginnings str = unwords $ map tag ch
where
tag x
| isPrefixOf "**" x = "<strong>" ++ (tail $ tail x)
| isPrefixOf "__" x = "<strong>" ++ (tail $ tail x)
| isPrefixOf "_" x = "<em>" ++ (tail x)
| isPrefixOf "*" x = "<em>" ++ (tail x)
| isPrefixOf "^" x = "<p>" ++ (tail x)
| isPrefixOf "---" x = replace "---" "<hr>"
| otherwise = x
ch = splitOn " " str
请记住，在Haskell中，the layout of your code matters。