Question

我想将我的json转换为以下格式。并将以下格式转换为我的记录。请检查我在下面写的代码。

{
    "uid" : "bob",
    "emailid" : "bob@bob.com",
    "email_verified" : "Y" // "Y" for EmailVerified and "N" for EmailNotVerified
}

我有以下代码，我试图将用户类型转换为json和从json转换用户类型在Haskell中使用Aeson库

{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE DeriveGeneric #-}

import Data.Monoid ((<>))
import GHC.Generics
import Data.Aeson (FromJSON, ToJSON)
import Data.Aeson.Types

data User = User {
    userId :: String,
    userEmail :: String,
    userEmailVerified :: EmailState
  } deriving (Show, Generic)

data EmailState = EmailVerified | EmailNotVerified deriving (Generic, Show)

instance ToJSON User where
    toJSON u = object [
       "uid" .= userId u,
       "emailid" .= userEmail u,
       "email_verified" .= userEmailVerified u
       ]

instance FromJSON User where
    parseJSON = withObject "User" $ \v -> User
        <$> v .: "uid"
        <*> v .: "emailid"
        <*> v .: "email_verified"

instance ToJSON EmailState
instance FromJSON EmailState

但是，我目前能够生成的格式如下所示

{
    "uid" : "bob",
    "emailid" : "bob@bob.com",
    "email_verified" : "EmailVerified"
}

Answer 1

您需要自己实现ToJSON EmailState（因此请删除instance ToJSON EmailState，并将其写为：）

instance ToJSON EmailState where
    toJSON EmailVerified = String "Y"
    toJSON EmailNotVerified = String "N"

您还需要更改解析器：

instance FromJSON EmailState where
    parseJSON (String "Y") = return EmailVerified
    parseJSON (String "N") = return EmailNotVerified
    parseJSON _ = fail "Invalid JSON value to parse"

Answer 2

威廉姆斯的令人敬畏的答案，只是为了添加语法您可以使用-

instance FromJSON EmailState where
    parseJSON (String s) 
        | s == "Y" = return EmailVerified
        | s == "N" = return EmailNotVerified
        | otherwise = fail "Invalid JSON value to parse"

-注意：否则= mzero-首选

ref：https://mail.haskell.org/pipermail/beginners/2011-October/008791.html

将记录中的构造函数转换为aeson haskell中的自定义json字符串

2 个答案: