mat*_*ias 2 generics json haskell custom-data-type aeson
如果我有自定义数据类型用于使用Aeson解析JSON
data Response = Response
{ response :: [Body]
} deriving (Show)
instance FromJSON Response where
parseJSON (Object v) = Response <$> v .: "response"
parseJSON _ = mzero
data Body = Body
{ body_id :: Int
, brandId :: Int
} deriving (Show)
instance FromJSON Body where
parseJSON (Object v) = Body
<$> v .: "id"
<*> v .: "brandId"
parseJSON _ = mzero
raw :: BS.ByteString
raw = "{\"response\":[{\"id\":5977,\"brandId\":87}]}"
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赠送:
?> decode raw :: Maybe Response
Just (Response {response = [Body {body_id = 5977, brandId = 87}]})
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如何FromJSON自动派生实例?
我试过了:
data Response = Response
{ response :: [Body]
} deriving (Show,Generic)
data Body = Body
{ body_id :: Int
, brandId :: Int
} deriving (Show,Generic)
instance FromJSON Response
instance FromJSON Body
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正如一些教程中所建议的那样,但这给出了:
?> :l response.hs
[1 of 1] Compiling Response ( response.hs, interpreted )
response.hs:19:22:
Can't make a derived instance of `Generic Response':
You need DeriveGeneric to derive an instance for this class
In the data declaration for `Response'
response.hs:24:22:
Can't make a derived instance of `Generic Body':
You need DeriveGeneric to derive an instance for this class
In the data declaration for `Body'
Failed, modules loaded: none.
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我究竟做错了什么?
Bak*_*riu 13
错误告诉您的是,您必须启用DeriveGeneric扩展才能使其正常工作.所以你必须添加:
{-# LANGUAGE DeriveGeneric #-}
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在文件的顶部,或使用-XDeriveGeneric标志进行编译.
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