继续我对管道和 aeson 的探索,我将如何使用我自己的数据类型,而不是使用Yesod bookValue
中的这个(稍微修改过的)代码片段。
{-# LANGUAGE OverloadedStrings, TemplateHaskell #-}
import Network.Wai (Response, responseLBS, Application, requestBody)
import Network.HTTP.Types (status200, status400)
import Network.Wai.Handler.Warp (run)
import Data.Aeson.Parser (json)
import Data.Conduit.Attoparsec (sinkParser)
import Control.Monad.IO.Class (liftIO)
import Data.Aeson (Value(..), encode, object, (.=))
import Control.Exception (SomeException)
import Data.ByteString (ByteString)
import Data.Conduit (ResourceT, ($$))
import Control.Exception.Lifted (handle)
import qualified Data.HashMap.Strict as M
import Data.Aeson.TH (deriveJSON)
-- I ADDED THIS
data JSONRequest = JSONRequest {
command :: ByteString,
params :: M.HashMap ByteString ByteString
}
deriveJSON id ''JSONRequest
-- END OF WHAT I ADDED
main :: IO ()
main = run 3000 app
app :: Application
app req = handle invalidJson $ do
value <- requestBody req $$ sinkParser json
newValue <- liftIO $ dispatch value
return $ responseLBS
status200
[("Content-Type", "application/json")]
$ encode newValue
invalidJson :: SomeException -> ResourceT IO Response
invalidJson ex = return $ responseLBS
status400
[("Content-Type", "application/json")]
$ encode $ object
[ ("message" .= show ex)
]
-- Application-specific logic would go here.
dispatch :: Value -> IO Value
dispatch = return
基本上,我想将类型更改dispatch
为 JSONRequest -> IO JSONRequest。如何告诉解析器使用我自己的 fromJSON 派生实例?
我尝试只添加一个类型声明,为 json 上的多态返回类型祈祷,但我意识到它严格用于 Value。