haskell – 使用包含保留关键字的字段名解析JSON

我试图用aeson解析下面的 JSON.

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

既然有很多字段我想忽略,似乎是I should use GHC generics.但是如何编写使用Haskell关键字(如数据和类型)的数据类型定义？以下当然给出：输入`data’的解析错误

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

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

解决方法

您可以编写自己的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.

帖子示例：

{-# 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"

饲料可以做同样的事情.如果不必忽略字段,那么也可以使用来自Data.Aeson.TH的derivedJSON,它可以修改字段名称作为第一个参数.