Haskell (Aeson) : How to construct a record from JSON with additional values

Question

I have a record that I want to construct with some values from a JSON string as well as some additionally provided values.

For example - given the following record:

data MyRecord = MyRecord { a :: String, b :: Int, c :: String }

I want do define a function:

createMyRecord :: String -> String -> Maybe MyRecord
createMyRecord json cValue = ???

Which I want to be able to call like so:

createMyRecord "{\"a\": \"a value\", \"b\": 100}" "c value"

Currently, I'm doing this by use Aeson to create the record with defaults (i.e. empty strings and zeros) for the values which don't come from the JSON. I then create a new version of the record with the other fields updated. Something like this:

instance FromJSON MyRecord where
    parseJSON = withObject "MyRecord" $ \o -> do
        valueA <- o .: "a"
        valueB <- o .: "b"
        return MyRecord { a = valueA, b = valueB, c = "" }

createMyRecord :: String -> String -> Maybe MyRecord
createMyRecord json cValue =
    Aeson.decode json <$> (\r -> r { c = cValue })

This feels a bit cumbersome - I'd like to create the record in one go with all the values, rather than filling them in step by step. Is there a nicer way to do this with Aeson (I'm open to other libraries also) that anyone can recommend?

Thanks!

Daniel Wagner · Accepted Answer · 2018-05-18 21:22:58Z

4

Do you want to build parseJSON? It doesn't have to be parseJSON...

abParser :: String -> Value -> Parser MyRecord
abParser valueC = withObject "MyRecord" $ \o -> do
    valueA <- o .: "a"
    valueB <- o .: "b"
    return MyRecord { a = valueA, b = valueB, c = valueC }

createMyRecord :: ByteString -> String -> Maybe MyRecord
createMyRecord json valueC = decode json >>= parseMaybe (abParser valueC)

edited May 18, 2018 at 21:22

answered May 18, 2018 at 21:09

Daniel Wagner

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1 Comment

tomphp Over a year ago

Excellent, thank you! I knew there was a simple solution but was struggling to work it out.

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