Parse JSON with fieldnames that contain reserved keywords

Question

I'm trying to parse the following JSON with aeson.

{
    "data": [
        {
            "id": "34",
            "type": "link",
            "story": "foo"
        },
        {
            "id": "35",
            "type": "link",
            "story": "bar"
        }
    ]
}

Since there are a lot of field I'd like to ignore, it seems I should use GHC generics. But how to write a data type definition that uses Haskell keywords like data and type? The following of course gives: parse error on input `data'

data Feed = Feed {data :: [Post]}
    deriving (Show, Generic)

data Post = Post {
        id :: String,
        type :: String,
        story :: String
    }
    deriving (Show, Generic)

Answer 1

You can write your own FromJSON and ToJSON instances without relying on GHC.Generics. This also means that you can use different field names for the data representation and the JSON representation.

Example instances for Post:

{-# LANGUAGE OverloadedStrings #-}
import Control.Applicative
import Data.Aeson
import qualified Data.ByteString.Lazy as LBS

data Post = Post {
        postId :: String,
        typ :: String,
        story :: String
  }
  deriving (Show)

instance FromJSON Post where
  parseJSON (Object x) = Post <$> x .: "id" <*> x.: "type" <*> x .: "story"
  parseJSON _ = fail "Expected an Object"

instance ToJSON Post where
  toJSON post = object 
    [ "id" .= postId post
    , "type" .= typ post
    , "story" .= story post
    ]

main :: IO ()
main = do
  print $ (decode $ Post "{\"type\": \"myType\", \"story\": \"Really interresting story\", \"id\" : \"SomeId\"}" :: Maybe Post)
  LBS.putStrLn $ encode $ Post "myId" "myType" "Some other story"

The same can be done for Feed. If you didn't have to ignore fields, you could also use deriveJSON from Data.Aeson.TH, which takes a function to modify field names as it's first argument.