I have the following JSON snippet:
{
"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
}
我希望能够解析以下类型的内容:
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实现,但是一直遇到错误。最终我成功让所有类型匹配上了,但解析出错了,我不太明白它在哪里出错了。
{-# 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
我只是得到了下面的输出,这并没有给我足够的信息来知道我的问题在哪里。
$ runhaskell lib/code.hs
Left "Error in $: expected main, encountered Object"
我已经将整个内容放在了一个Gist中,可以在这里查看。
我想知道代码有什么问题,以及如何修复它。如果你有关于如何更易读地编写此代码的建议,我也很想知道。目前,我主要对两个独立的函数
lambda3
和parseInner
感到恼火。
V.head
,您可以获取weather
数组中的第一个元素,该元素是一个对象,因此应为lambda3 = withObject "weatherMain" (.: "main")
。 - zakyggaps