使用包含保留关键字的字段名解析JSON

Question

我正在尝试用aeson解析以下JSON.

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

由于我想忽略很多领域,似乎我应该使用GHC泛型.但如何编写使用哈斯克尔关键字等的数据类型定义data和type？以下当然给出: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

你可以写自己FromJSON和ToJSON实例,而不依赖于GHC.Generics.这也意味着您可以为数据表示和JSON表示使用不同的字段名称.

Post的示例实例:

{-# 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"

Feed也可以这样做.如果你不必忽略字段,你也可以使用deriveJSONfrom Data.Aeson.TH,它接受一个函数来修改字段名称作为它的第一个参数.