Haskell:在没有数据的字符串中替换子字符串.包列表

Question

我是Haskell的新手，我想知道如何用另一个单词替换字符串中的预定单词。这是我到目前为止的代码，我知道我现在不能这样做:

treat :: String -> String
treat text = text
main::IO()
main = do argv <- getArgs
texte <- readFile "intputText"
print (separation text)
print ( treat text )

separation :: String -> [String]
separation [] = [""]
separation (c:cs) | c == "Graph"  = "Graphic : " : rest
| c == '}'  = "" : rest
| c == '{'  = "" : rest
| otherwise = (c : head rest) : tail rest
where rest = separation cs

所以基本上我知道我不能在第一个c == "Graph"所以我想知道我基本上可以把每个单词"Graph"in my Stringtextby "Graphic"

我希望能够做到这一点，而不需要导入任何包。

如果有人能帮我，我将非常感激。

非常感谢!

Answer 1

replace :: String -> String -> String-> String
replace [] token repl = []
replace str@(s:ss) token@(t:tx) repl
-- check if first char of string equal to first char of token
| s == t = case validateToken token str of
Just list -> repl ++ replace list token repl
Nothing -> s : replace ss token repl
-- if not equal then continue recursion step
| otherwise = s: replace ss token repl
where
-- validate if token matches the following chars of the string
-- returns Nothing if token is not matched
-- returns the remaining string after the token if token is matched
validateToken:: String -> String -> Maybe String
validateToken (a:as) [] = Nothing
validateToken [] list = Just list
validateToken (a:as) (x:xs) 
| a == x = validateToken as xs 
| otherwise = Nothing
example = replace "yourString" "token" "new"