Json Haskell-Aeson解析混合元素数组
我有一个类似的json:Json Haskell-Aeson解析混合元素数组,json,parsing,haskell,aeson,Json,Parsing,Haskell,Aeson,我有一个类似的json: { "name" : "Sam", "items": [ "sword", "shield", [] ] } 和数据类型 data Adventurer = Adventurer { name :: String, items :: [String] } deriving (Generic, Show, FromJSON, ToJSON) 问题的原因是“items”字段在数组中有额外的[],而我得到的是“expected String,expecte
{
"name" : "Sam",
"items": [ "sword", "shield", [] ]
}
和数据类型
data Adventurer = Adventurer {
name :: String,
items :: [String]
} deriving (Generic, Show, FromJSON, ToJSON)
问题的原因是“items”字段在数组中有额外的[],而我得到的是“expected String,expected array”
我一直在尝试使用“items”数组元素的自定义数据类型来解决这个问题,或者尝试创建一个自定义解析器来忽略额外的数组
有没有一种方法可以解析一个数组,只取某一类型的项,而不取其他项?是的,例如,我们可以首先构造一个函数,将
值
s(这些是JSON对象)映射到可能的字符串
s:
import Data.Aeson(Value(String))
import Data.Text(unpack)
getString :: Value -> Maybe String
getString (String t) = Just (unpack t)
getString _ = Nothing
然后,我们可以为Adventurer
定义FromJOSN
的自定义实现,如:
{-# LANGUAGE OverloadedStrings #-}
import Data.Aeson(FromJSON(parseJSON), withObject, (.:))
import Data.Maybe(catMaybes)
instance FromJSON Adventurer where
parseJSON = withObject "Adventurer" $ \v -> Adventurer
<$> v .: "name"
<*> fmap (catMaybes . map getString) (v .: "items")
Main> t = "{\n \"name\" : \"Sam\",\n \"items\": [ \"sword\", \"shield\", [] ]\n}"
Main> decode t :: Maybe Adventurer
Just (Adventurer {name = "Sam", items = ["sword","shield"]})