Aeson编码嵌套记录

Question

假设有这样的记录：

data Place = Place
  { details :: PlaceDetails
  , amenities :: [Amenity]
  , photos :: [PlacePhoto]
  } deriving (Generic)

现在，它是这样的：

instance ToJSON Place where
  toEncoding Place{..} = 
    pairs  $ "details" .= details
          <> "amenities" .= amenities
          <> "photos" .= photos

我使用toEncoding来保存我想要的特定顺序。但是，我想从编码中删除details键，以便PlaceDetails中所有字段与amenities和photos处于同一级别，而无需手动指定所有这些字段，因为PlaceDetails非常大。我该怎么做呢？我知道我可以用HML.unions合并[Value]，但我只用toJSON做了这件事，而不是toEncoding。

我目前所拥有的：

{
    "details": {
        "id": "place_6de6cda0f8524a6f9c264c84afdbadad",
        "name":"somename",
        "description":"some description", 
        "other": "a lot more fields"
    }
    "amenities": []
    "photos": []
}

这就是我想要的：

{
    "id":"place_6de6cda0f8524a6f9c264c84afdbadad",
    "name":"somename",
    "description":"some description", 
    "other": "a lot more fields",
    "amenities": []
    "photos": []
}

谢谢。

Answer 1

一个Aeson对象只是一个哈希图，你可以使用<> (中缀)或mappend将其联合起来。

例如：

#!/usr/bin/env cabal
{- cabal:
    build-depends: base, aeson
-}
{-# LANGUAGE DeriveAnyClass    #-}
{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE OverloadedLists   #-}
{-# LANGUAGE DeriveGeneric     #-}
{-# LANGUAGE RecordWildCards   #-}

module Main where

import Data.Aeson
import GHC.Generics

main :: IO ()
main = print (encode (Place (Foo 1 2) (Bar 3 4)))

data Foo = Foo {foo1 :: Int, foo2 :: Int}
    deriving (Eq,Ord,Show,Generic,ToJSON)

data Bar = Bar {bar1 :: Int, bar2 :: Int}
    deriving (Eq,Ord,Show,Generic,ToJSON)

data Place = Place { foo :: Foo , bar :: Bar }
    deriving (Eq,Ord,Show,Generic)

instance ToJSON Place where
  toJSON Place{..} =
       let Object b = toJSON bar
       in Object ([("foo",toJSON foo)] <> b)

我们可以运行下面的代码：

% chmod +x so.hs && ./so.hs
... snippet out build info ...
"{\"bar1\":3,\"foo\":{\"foo2\":2,\"foo1\":1},\"bar2\":4}"