假设你有一个简单的类型,并且想用Aeson将它序列化为JSON。以下是基本设置:如何在Haskell中将参数化类型序列化为JSON?
{-# LANGUAGE DeriveGeneriC#-}
import Data.Aeson (ToJSON)
import Data.Aeson (encode)
import GHC.Generics
data Spotting = Spotting {
state :: String,
bird :: String
} deriving (Show, Generic)
instance ToJSON Spotting
现在说你想要的,除了bird
和state
领域,允许用户在加时赛/自定义元数据传递。对于鸟类发现,也许这是气温,鸟类的密度,潮汐的位置......可能是任何事情,我们事先不知道。
从看着像Twitter API in Haskell的例子,它看起来像你想构建这样的:
data Spotting meta = Spotting {
state :: String,
bird :: String,
meta :: meta
} deriving (Show, Generic)
instance ToJSON meta => ToJSON (Spotting meta)
从我的理解,到目前为止,这是一个参数化类型。目标是现在做一个简单的方法来创建一些JSON。所以我们定义一个这样的函数:
spotting bird state meta = encode $ Spotting {
state = state,
bird = bird,
meta = meta
}
但是我不知道该从哪里走。当我这样调用该函数:
record = spotting "Snowy Egret" "California" "low tide"
它抛出一个错误(我是新来的Haskell所以我仍然在学习如何解释这一切东西的基础知识)
No instance for (Data.String.IsString meta0)
arising from the literal `"low tide"'
The type variable `meta0' is ambiguous
Possible fix: add a type signature that fixes these type variable(s)
Note: there are several potential instances:
instance Data.String.IsString
aeson-0.7.0.6:Data.Aeson.Types.Internal.Value
-- Defined in `aeson-0.7.0.6:Data.Aeson.Types.Internal'
instance a ~ Data.ByteString.Internal.ByteString =>
Data.String.IsString
(attoparsec-0.12.1.2:Data.Attoparsec.ByteString.Internal.Parser a)
-- Defined in `Data.Attoparsec.ByteString.Char8'
instance Data.String.IsString Data.Text.Internal.Text
-- Defined in `Data.Text'
...plus five others
In the third argument of `spotting', namely `"low tide"'
In the expression:
spotting "Snowy Egret" "California" "low tide"
In an equation for `record':
record = spotting "Snowy Egret" "California" "low tide"
发生了什么在这里/你如何得到这个工作?
的最终目标是,不要在meta
场传递一个字符串,而是一个类型的对象(但它可以是任何对象),像这样的:
record = spotting "Snowy Egret" "California" MyCustomData {
tide = "low"
}
你是怎么做到这一点哈斯克尔?
'可能的修复:添加一个修复这些类型变量的类型签名' – alternative 2014-11-02 23:20:25