如何在haskell中用另一个替换字符串

Veg*_*512 19 haskell

我想用不同的字符串替换输入文件中的字符串.我正在寻找一种方法,但似乎我只能按字符改变字符串.例如,在下面的代码中

replace :: String -> String 
replace [] = [] 
replace (x:xs) = if x == '@' then 'y':replace xs --y is just a random char
                             else x:replace xs

searching :: String -> IO String
searching filename = do
    text <- readFile filename
    return(replace text)


main :: IO ()
main = do

  n <- searching "test.sf"
  writeFile "writefile.html" n 
Run Code Online (Sandbox Code Playgroud)

我想找到第一次出现的字符串"@title",但我似乎找不到如前所述的方法,我只能访问字符'@'.有没有办法完成这样的任务.

jos*_*uan 21

您可以使用Data.List.Utils替换,它是懒惰的,您可以处理一个大文件,如:

main = getContents >>= putStr . replace "sourceString" "destinationString"
Run Code Online (Sandbox Code Playgroud)

就这样!

可能的替换功能可能是

rep a b s@(x:xs) = if isPrefixOf a s

                     -- then, write 'b' and replace jumping 'a' substring
                     then b++rep a b (drop (length a) s)

                     -- then, write 'x' char and try to replace tail string
                     else x:rep a b xs

rep _ _ [] = []
Run Code Online (Sandbox Code Playgroud)

另一种聪明的方式(来自Data.String.Utils)

replace :: Eq a => [a] -> [a] -> [a] -> [a]
replace old new l = join new . split old $ l
Run Code Online (Sandbox Code Playgroud)

  • 或者,使用`split`包中的[Data.List.Split](http://hackage.haskell.org/packages/archive/split/0.2.1.1/doc/html/Data-List-Split.html)是Haskell平台的一部分,定义`replace old new = intercalate new.splitOn old`. (12认同)