2017-07-03 57 views
0

我有一個JSON結構這樣埃宋:解析動態結構

{ 
    "tag1": 1, 
    "tag2": 7, 
    ... 
} 

而且我有一種這樣的

data TagResult { name :: String, numberOfDevicesTagged :: Int } deriving (Show, Eq) 
newtype TagResultList = TagResultList { tags :: [TagResult] } 

標籤名稱是當然的完全動態的,我不知道他們在編譯時。 我想創建一個實例FromJSON來解析JSON數據,但我無法編譯它。如何定義parseJSON以實現此目的?

+0

您可以對'Map'使用現有的'FromJSON'和'ToJSON'類型。您的標籤將成爲關鍵。 –

+1

[FromJSON可能出現多個字段列表](https://stackoverflow.com/questions/44514645/fromjson-make-a-list-from-multiple-fields) –

+0

這是與[FromJSON make來自多個字段的列表](https://stackoverflow.com/questions/44514645/fromjson-make-a-list-from-multiple-fields)。前面提到的鏈接適用於編譯時已知可能標記列表的情況,在這種情況下它們是未知的。 – Batou99

回答

1

您可以使用Object是HasMap並在運行時提取該鍵的事實。然後,您可以按如下方式編寫FromJSON實例 -

{-# LANGUAGE OverloadedStrings #-} 
module Main where 

import Data.Aeson 
import qualified Data.Text as T 
import qualified Data.HashMap.Lazy as HashMap 

data TagResult = TagResult { name :: String 
          , numberOfDevicesTagged :: Int 
          } deriving (Show, Eq) 


newtype TagResultList = TagResultList { tags :: [TagResult] } deriving Show 


instance ToJSON TagResult where 

    toJSON (TagResult tag ntag) = 
    object [ T.pack tag .= ntag ] 

instance ToJSON TagResultList where 

    toJSON (TagResultList tags) = 
    object [ "tagresults" .= toJSON tags ] 


instance FromJSON TagResult where 

    parseJSON (Object v) = 
    let (k, _) = head (HashMap.toList v) 
    in TagResult (T.unpack k) <$> v .: k 

    parseJSON _ = fail "Invalid JSON type" 

instance FromJSON TagResultList where 

    parseJSON (Object v) = 
    TagResultList <$> v .: "tagresults" 


main :: IO() 
main = do 

    let tag1 = TagResult "tag1" 1 
     tag2 = TagResult "tag2" 7 
     taglist = TagResultList [tag1, tag2] 

    let encoded = encode taglist 
     decoded = decode encoded :: Maybe TagResultList 

    print decoded 

上述程序應打印標籤結果列表。

Just (TagResultList {tags = [TagResult {name = "tag1", numberOfDevicesTagged = 1},TagResult {name = "tag2", numberOfDevicesTagged = 7}]})