如何正確地爲我的數據類型實現FromJSON？

Question

我有幾個構造一個數據：

data MyData = 
    A1 { var1 :: Int, var2 :: String } 
    A2 { var1 :: Int, var2 :: String, var3 :: Char } 
    A3 { var1 :: Int, var2 :: Char, var3 :: String, var4 :: String } 

這是我真正的「數據」的簡化版本。在沒有使用deriving (Generic)的情況下，我怎樣才能實現它爲ToJSON類？

instance FromJSON MyData where 
    parseJSON (Object v) = do 
    .. --- how do I know if v is created by A1 or A2 or A3? 

Answer 1

埃宋的Parser是Alternative一個實例。因此，您可以提供多個解析器和與(<|>)將它們結合起來：

{-# LANGUAGE OverloadedStrings #-} 
import Control.Applicative ((<|>)) 
import Data.Aeson 

-- Removed record fields to keep things short, but you can put them back. 
data MyData = A1 Int String 
      | A2 Int String Char 
      | A3 Int Char String String 
      deriving Show 

instance FromJSON MyData where 
    parseJSON (Object v) = A3 <$> v .: "var1" <*> v .: "var2" <*> v .: "var3" <*> v .: "var4" 
         <|> A2 <$> v .: "var1" <*> v .: "var2" <*> v .: "var3" 
         <|> A1 <$> v .: "var1" <*> v .: "var2" 

注意解析器的順序很重要，因爲v .: "var1"和v .: "var2"所有三個解析器成功。