Haskell string replace - string

Question

How do I replace a character with multiple characters in a string.
I can replace a character with another character with below:
replaceO [] = []
replaceO (x:xs) =
if x == '('
then '\n' : replaceO xs
else x : replaceO xs
But I want to replace '(' with "\n\t\t", not just '\n'.
---------- UPDATE --------------
Based on the answer below, can I make that a function that accepts a string and returns a string like this:
ReplaceFun str :: String -> String
ReplaceFun str = do
concatMap (\str -> if str == '.' then "foo" else [str])
This doesn't work, can someone point out my mistake? I am very new to Haskell.
Here is what I have for nested indentation:
replaceO (x:xs) n l =
if x == '('
then "\n" ++ (showTabs n "-") ++ replaceO xs (n + 1) 'l'
else
if x == ')'
then "\n" ++ (showTabs n "-") ++ replaceO xs (n - 1) 'l'
else x : replaceO xs n 'l'

Answer 1

Just use concatMap
Prelude> concatMap (\x -> if x == '.' then "foo" else [x]) "example..."
"examplefoofoofoo"

Answer 2

expanding your example, you can substitute "\n\t\t" for "(" by just prepending multiple things to the result of the recursive call
replace1 [] = []
replace1 (x:xs) =
if x == '('
then '\n' : '\t' : '\t' : replace1 xs
else x : replace1 xs
of course, this is equivalent to using "\n\t\t" ++
replace2 [] = []
replace2 (x:xs) =
if x == '('
then "\n\t\t" ++ replace2 xs
else x : replace2 xs
if we note that x : is equivalent to [x] ++
replace3 [] = []
replace3 (x:xs) =
if x == '('
then "\n\t\t" ++ replace3 xs
else [x] ++ replace3 xs
then we can factor out the repeated recursive call
replace4 [] = []
replace4 (x:xs) = (if x == '(' then "\n\t\t" else [x]) ++ replace4 xs
then for clarity, we could filter out the if statement into a function:
replace5 [] = []
replace5 (x:xs) = f x ++ replace5 xs
where f x = if x == '('
then "\n\t\t"
else [x]
we could then reorganize our code - rather than alternately applying f to each x and
appending it to our results, we could apply f to all the xs and then concatenate all the results:
replace6 xs = concat $ map f xs
where f x = if x == '('
then "\n\t\t"
else [x]
But concat $ map f xs is used so often that it has another name:
replace7 xs = concatMap f xs
where f x = if x == '('
then "\n\t\t"
else [x]
Actually, it has another one too, because [] is a monad:
replace8 xs = xs >>= f
where f x = if x == '('
then "\n\t\t"
else [x]
But if we're doing that, we might as well go full Monad:
replace9 xs = do
x <- xs
if x == '('
then "\n\t\t"
else [x]
But I'd probably just stop at replace7, personally.