如何用haskell中的另一个替换字符串 [英] How to replace a string with another in haskell

问题描述

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，但似乎我只能改变字符的字符串。例如在下面的代码中：

  replace :: String  - >字符串
替换[] = [] 
替换（x：xs）= if x =='@'然后'y'：替换xs --y只是一个随机字符
 else x ：替换xs 
 
 searching :: String  - > IO字符串
搜索文件名= do 
文本<  -  readFile文件名称
返回（替换文本）
 
 
 main :: IO（）
 main = do 
 
n<  - 搜索test.sf
 writeFilewritefile.htmln 
  

我想查找字符串@title的第一个匹配项，但我似乎找不到前面提到的方法，我只能访问char'@' 。有没有一种方法来完成这样的任务。

解决方案

您可以使用> Data.List.Utils 替换，它很懒，你可以用一些像处理一个大文件：

  main = getContents>> = putStr。替换sourceStringdestinationString
  

可能的替换函数可能是

  rep abs @（x：xs）= if isPrefixOf as 
 
  - 然后写'b'并替换跳跃的'a'子串
然后b ++ rep ab（drop（length a）s）
 
  - 然后写'x'char并尝试替换尾字符串
 else x：rep ab xs 
 
 rep _ _ [] = [] 
  

另一个聪明的方式（来自Data.String.Utils）

  replace :: Eq a => [a]  - > [a]  - > [a]  - > [a] 
替换旧的新的l =加入新的。拆分旧$ l 
  

I want to replace a string from an input file with a different string. I was searching for a method but it seems i can only alter the string character by character. For example in the my code below

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 

I want to find the first occurrence of the string "@title", but i cant seem to find a method to do so as mentioned before, i can only access the char '@'. Is there a method for doing such a task.

解决方案

You can use Data.List.Utils replace, it's lazy and you can process a big file with some like:

main = getContents >>= putStr . replace "sourceString" "destinationString"

That's all!

A possible replace function could be

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 _ _ [] = []

another smart way (from Data.String.Utils)

replace :: Eq a => [a] -> [a] -> [a] -> [a]
replace old new l = join new . split old $ l

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