Find all capturing groups of a regular expression

Question

I am looking for a Haskell function that returns the capturing groups of all matches of a given regex.

I have been looking at Text.Regex, but couldn't find anything there.

Now I am using this workaround which seems to work:

import Text.Regex

findNext :: String -> Maybe (String, String, String, [String] ) -> [ [String] ]
findNext pattern Nothing = []
findNext pattern (Just (_, _, rest, matches) ) = 
    case matches of
        [] -> (findNext pattern res)
        _ -> [matches] ++ (findNext pattern res)
    where res = matchRegexAll (mkRegex pattern) rest

findAll :: String -> String -> [ [String] ]
findAll pattern str = findNext pattern (Just ("", "", str, [] ) )

Result:

findAll "x(.)x(.)" "aaaxAxaaaxBxaaaxCx"
[["A","a"],["B","a"]]

Question:

• Did I miss something in Text.Regex?
• Is there a Haskell regex library that implements a findAll function?

Answer 1

You can use the =~ operator from Text.Regex.Posix:

Prelude> :mod + Text.Regex.Posix
Prelude Text.Regex.Posix> "aaaxAxaaaxBxaaaxCx" =~ "x(.)x(.)" :: [[String]]
[["xAxa","A","a"],["xBxa","B","a"]]

Note the explicit [[String]] type. Try replacing it with Bool, Int, String and see what happens. All types that you can use in this context are listed here. Also see this tutorial.