Question

我有以下JSON代码段：

{
  "weather": [
    {
      "id": 803,
      "main": "Clouds",
      "description": "broken clouds",
      "icon": "04n"
    }
  ],
  "main": {
    "temp": 271.979,
    "pressure": 1024.8,
    "humidity": 100,
    "temp_min": 271.979,
    "temp_max": 271.979,
    "sea_level": 1028.51,
    "grnd_level": 1024.8
  },
  "id": 6332485,
  "name": "Queensbridge Houses",
  "cod": 200
}

我想从中解析以下类型：

data WeatherResponse = WeatherResponse
  { temp :: Double
  , humidity :: Double
  , weatherMain :: T.Text
  } deriving Show

我一直在尝试使用以下代码来执行此操作，但我一直遇到错误。我终于得到了所有类型的匹配，但它解析不正确，并且不能真正理解它失败的地方。

{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE RecordWildCards #-}
{-# LANGUAGE ScopedTypeVariables #-}

import Data.Aeson
import Data.Aeson.Types (Parser, Array)
import Data.Time (defaultTimeLocale, formatTime, getZonedTime)

import qualified Data.ByteString.Lazy as BL
import qualified Data.Vector as V
import qualified Data.Text as T

data WeatherResponse = WeatherResponse
  { temp :: Double
  , humidity :: Double
  , weatherMain :: T.Text
  } deriving Show

lambda3 :: Value -> Parser T.Text
lambda3 o = do
  withText "main" (\t -> do
                      return t
                  ) o

parseInner :: Value -> Parser T.Text
parseInner a = withArray "Inner Array" (lambda3 . (V.head)) a

instance FromJSON WeatherResponse where
  parseJSON =
    withObject "Root Object" $ \o -> do
    mainO <- o .: "main"
    temp <- mainO .: "temp"
    humidity <- mainO .: "humidity"
    weatherO <- o .: "weather"
    weatherMain <- parseInner weatherO
    return $ WeatherResponse temp humidity weatherMain

getSampleData = BL.readFile "/home/vmadiath/.xmonad/weather.json"

main = do
  text <- getSampleData
  let (result :: Either String WeatherResponse) = eitherDecode text
  putStrLn . show  $ result

我只是得到以下输出，这些输出并不足以让我知道我哪里出错了。

$ runhaskell lib/code.hs
Left "Error in $: expected main, encountered Object"

我已将整件事放在一个可见的here

中

我想知道代码有什么问题，以及我如何解决它。如果你有一个如何以更易读的方式写这个的建议，我也很想知道。目前，我对两个独立的函数lambda3和parseInner感到烦恼而烦恼）

Answer 1

在我看来，你这太复杂了。这样的事情应该有效：

instance FromJSON WeatherResponse where
  parseJSON (Object v) = do
      weatherValue <- head <$> v .: "weather"
      WeatherResponse <$> ((v .: "main") >>= (.: "temp"))
                          <*> ((v .: "main") >>= (.: "humidity"))
                          <*> weatherValue .: "main"

它是输出：

[nix-shell:~/haskell-sample]$ ./weather
Right (WeatherResponse {temp = 271.979, humidity = 100.0, weatherMain = "Clouds"})

如何使用Aeson解析此JSON？

1 个答案: