How can you modify a function for your own type?

Question

Haskell

simple types and a complex type structure are defined , the following functions should have been implemented for them ... in particular, I had a problem with the latter: a list of all identifiers with a numeric type).
when you enter getByTypes [MyInt, MyString] base returns
["first","next","str"] , but if you set MyStruct as an argument, it gives an error, how to make the function work with this type?

data MyType = Null |
              MyInt |
              MyDouble |
              MyString |
              MyStruct [(String,MyType)] deriving (Show,Eq)

isStructured :: MyType -> Bool
isStructured (MyStruct a) = True
isStructured (_) = False

getType :: String -> [(String, MyType)] -> MyType
getType _ [] = Null
getType s (h:t)  | fst(h) == s = snd(h)
                 | otherwise = getType s t

getFieldsFromStruct :: MyType -> [(String,MyType)]
getFieldsFromStruct (MyStruct a) = a
getFieldsFromStruct _ = []

getFields :: String -> [(String,MyType)] -> [(String,MyType)]
getFields _ [] = []
getFields s (h:t) | fst(h) == s && isStructured (snd h) = getFieldsFromStruct (snd h)
                  | otherwise = getFields s t

getByType :: MyType -> [(String,MyType)] -> [String]
getByType _ [] = []
getByType a (h:t)  | a == (snd h) = (fst h) : getByType a t
                   | otherwise = getByType a t

getByTypes :: [MyType] -> [(String,MyType)] -> [String]
getByTypes _ [] = []
getByTypes [] _ = []
getByTypes (h:t) a = (getByType h a) ++ getByTypes t a

base = [("first",MyInt), ("second",MyDouble), ("str",MyString), ("next",MyInt) ,("s",MyStruct[("s1",MyInt),("s2",MyDouble)])]

Answer 1

when you type getByTypes [MyInt, MyString] base returns ["first","next","str"]

So the function is working.

if you set MyStruct as an argument, it gives an error, how to make the function work with this type?

How is the call going?

getByTypes [MyStruct[("s1",MyInt),("s2",MyDouble)]] base