无法将类型“[Char]”与 spa 的“Char”匹配

Question

I am trying to give the names of spa a given supervisor that rated the skill service along with their rating result for each spa. But I keep getting this error shown below. Sorry if my code look horrible since I am new to Haskell.

* Couldn't match type `[Char]' with `Char'
  Expected type: Spa -> Service
    Actual type: Spa -> [Service]
* In the second argument of `(.)', namely `getService'
  In the first argument of `filter', namely `((== se) . getService)'
  In the expression: filter ((== se) . getService)

I tried by retrieving and filtering for the supervisor. Then I output it all. Is it possible to just output name of supervisor, spa and level rating? Am I going the right track? How to fix the error?

data Spa = Spa SID Brand Area Stars [(Service, LevelRating)]

-- spa service
getService :: Spa -> [Service]
getService (Spa _ _ _ _ xs) = map fst xs

-- filter spa service
spaService :: Service -> [Spa] -> [Spa]
spaService se = filter ((==se) . getService)

ratedListStr :: Spa -> String
ratedListStr (Spa sid br ar st s) = "\nSpaID: " ++ sid ++ "\n Brand: " ++ br ++ "\n Area: " ++ ar ++ "\n Star: " ++ show st ++ "\n Service Rating" ++ show s 

Answer 1

getService returns a list of Service s, zo [Service] , wheras se is a single Service , so (==se) does not make much sense. Since of of the operands is a Service and the other one is a [Service] . Since (==):: Eq a => a -> a -> Bool acts on two items of the same type, this does not typecheck.

If you want to check if the se is for example an element of the services, you can make use of elem:: Foldable f => a -> fa -> Bool :

spaService :: Service -> [Spa] -> [Spa]
spaService se = filter ( . getService)