2017-07-03 52 views
0

我有一个JSON结构这样埃宋:解析动态结构

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

而且我有一种这样的

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

标签名称是当然的完全动态的,我不知道他们在编译时。 我想创建一个实例FromJSON来解析JSON数据,但我无法编译它。如何定义parseJSON以实现此目的?

+0

您可以对'Map'使用现有的'FromJSON'和'ToJSON'类型。您的标签将成为关键。 –

+1

[FromJSON可能出现多个字段列表](https://stackoverflow.com/questions/44514645/fromjson-make-a-list-from-multiple-fields) –

+0

这是与[FromJSON make来自多个字段的列表](https://stackoverflow.com/questions/44514645/fromjson-make-a-list-from-multiple-fields)。前面提到的链接适用于编译时已知可能标记列表的情况,在这种情况下它们是未知的。 – Batou99

回答

1

您可以使用Object是HasMap并在运行时提取该键的事实。然后,您可以按如下方式编写FromJSON实例 -

{-# 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 

上述程序应打印标签结果列表。

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