如何从Json file.Json 结构中提取信息必须通过类型映射并使用AESON

Question

{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE DeriveGeneric #-}

module Main where

import           Data.Aeson
import           Data.Aeson.Encode.Pretty
import           Data.Aeson.Types
import           Data.ByteString as B
import qualified Data.ByteString.Lazy.Char8 as LC
import           GHC.Generics
import           Data.Text

data MainWeatherInfo = MainWeatherInfo
    { mainInfo :: Main
    } deriving (Show,Generic)

data Main = Main
    { temp :: Double
    , pressure :: Int
    , humidity :: Int
    , temp_min :: Double
    , temp_max :: Double
    } deriving (Show,Generic)

instance FromJSON MainWeatherInfo
instance FromJSON Main

main :: IO ()
main = do
  jsonString <- B.readFile "/home/ashot/test.json"
  let result = decodeStrict $ jsonString :: Maybe MainWeatherInfo
  case result of
    Nothing      -> print "error"
    Just result  -> print result

和JSON文件：

{
    "coord": {
        "lon": 44.51,
        "lat": 40.18
    },
    "weather": [{
        "id": 500,
        "main": "Rain",
        "description": "light rain",
        "icon": "10d"
    }],
    "base": "stations",
    "main": {
        "temp": 299.15,
        "pressure": 1008,
        "humidity": 34,
        "temp_min": 299.15,
        "temp_max": 299.15
    },
    "visibility": 10000,
        "wind": {
        "speed": 1.5,
        "deg": 190
    },
    "rain": {
        "3h": 0.195
    },
    "clouds": {
        "all": 8
    },
    "dt": 1528615800,
    "sys": {
        "type": 1,
        "message": 0.0036,
        "country": "AM",
        "sunrise": 1528594331,
        "sunset": 1528648255
    },
    "id": 616052,
    "name": "Yerevan",
    "cod": 200
}

我需要 Extract "main" info.And 我在启动时输入 MainWeatherInfo.This 代码是 compiled.But 我得到一个 "error"。我该如何解决这个问题并获得正常信息？

Answer 1

问题是 aeson 正在文件中寻找一个名为 mainInfo 的字段，但这样的字段不存在。

最初我认为我们可以通过将 MainWeatherInfo 的类型定义中的 mainInfo 更改为 main[ 来解决问题=43=]。这是行不通的，因为 已经存在同名函数！

aeson 提供了解决此问题的方法，您可以在 泛型：自定义字段名称[=43] 部分下找到 this helpful tutorial 中的说明=].

更具体地说，如果您使用 turorial 的方法，您最终会得到一个 Main.hs，如下所示：

{-# LANGUAGE OverloadedStrings #-} {-# LANGUAGE DeriveGeneric #-} {-# LANGUAGE DeriveAnyClass #-} module Main where import GHC.Generics import Data.Aeson import Data.Aeson.Types import qualified Data.ByteString as L data MainWeatherInfo = MainWeatherInfo { mainInfo :: Main } deriving ( Show , Generic ) instance ToJSON MainWeatherInfo where toJSON = genericToJSON defaultOptions { fieldLabelModifier = take 4 } instance FromJSON MainWeatherInfo where parseJSON = genericParseJSON defaultOptions { fieldLabelModifier = take 4 } data Main = Main { temp :: Double , pressure :: Int , humidity :: Int , temp_min :: Double , temp_max :: Double } deriving ( Show , Generic , ToJSON , FromJSON ) main :: IO () main = do contents <- L.readFile "notes/test.json" let result = (decodeStrict contents) :: Maybe MainWeatherInfo case result of Nothing -> putStrLn "error" Just v -> putStrLn $ show v

我在我的电脑上试了一下，效果很好。

更新： 我一直非常愚蠢并认为解决方案需要仅限于一个模块。有一个看起来像 Main.hs 的 much 可能会更好：

module Main where import Data.Aeson import qualified Data.ByteString as L import qualified Weather as W main :: IO () main = do contents <- L.readFile "notes/test.json" let result = (decodeStrict contents) :: Maybe W.MainWeatherInfo case result of Nothing -> putStrLn "error" Just v -> putStrLn $ show v

还有另一个模块 Weather.hs 即：

{-# LANGUAGE DeriveGeneric #-} {-# LANGUAGE DeriveAnyClass #-} module Weather where import GHC.Generics import Data.Aeson import Data.Aeson.Types data MainWeatherInfo = MainWeatherInfo { main :: Main } deriving ( Show , Generic , ToJSON , FromJSON ) data Main = Main { temp :: Double , pressure :: Int , humidity :: Int , temp_min :: Double , temp_max :: Double } deriving ( Show , Generic , ToJSON , FromJSON )

我测试了它只是为了确定，它确实有效。

如何从Json file.Json 结构中提取信息必须通过类型映射并使用AESON

How to extract information from Json file.Json structure must be mapped through type and use AESON

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haskell

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