如何统一提取Haskell记录中的字段名称和值 [英] How to generically extract field names and values in Haskell records

问题描述

$ b

  { - ＃LANGUAGE DeriveDataTypeable＃ - } $ b $我最近得知我可以在Haskell中执行以下操作： b 
 import Data.Data 
 
 data MyRecord = MyRecord 
 {field1 :: Int 
，field2 :: String 
，field3 :: String 
 
 $ main = print $ constrFields（toConstr（MyRecord 5HelloWorld））
  
 
 
 
这会给我以下信息：
  [ field1，field2，field3] 
  
 我该怎么做对于记录中的值，如下所示： 
  [5，Hello，World ] 
  
我在问，因为我正在使用 Aeson  
 $ p $ {
field1：5，
 code> field2：Hello，
field3：World
} 
  

然后像这样生成Haskell代码：

  field1 :: Int 
 field1 = 5 
 
 field2 :: String 
 field2 =Hello
 
 field3 :: String 
 field3 =世界
  

我如何一般地解开给定记录中的所有值，就像我可以解开我的记录的字段名称？

解决方案

您的第一个问题可以回答。如果您很乐意将数据类型的所有值转换为字符串，那么您确实可以使用通用编程库（例如， generics-sop ：

 { - ＃LANGUAGE DeriveGeneric，FlexibleContexts＃ - } 
 
将合格的GHC.Generics导入为G 
 import Generics.SOP 
 
数据MyRecord = MyRecord 
 {field1 :: Int 
，field2 :: String 
，field3 :: String 
}派生（Show，Eq，G.Generic）
 
实例Generic MyRecord 
 
 stringValues :: 
（Generic a，All2 Show（Code a））
 => a  - > [String] 
 stringValues a = 
 hcollapse（hcmap（Proxy :: Proxy Show）（\（I x） - > K（show x））（from a））
 
 test :: [String] 
 test = stringValues（MyRecord 5Helloworld）
  

现在在GHCi：

  GHCi>测试
 [5，\Hello \，\world \] 
  

$ b但是，如果您的目标是维护原始类型，那么这会更困难，因为结果类型必须是异类列表（实际上泛型 - sop 在使用 hcollapse ）转换回正常列表之前在内部使用。

我不完全清楚你真的想达到什么目的。很可能有一个更直接的解决方案。

I recently learned that I can do the following in Haskell:

{-# LANGUAGE DeriveDataTypeable #-}

import Data.Data

data MyRecord = MyRecord
  { field1 :: Int
  , field2 :: String
  , field3 :: String
  } deriving (Show,Eq,Data,Typeable)

main = print $ constrFields (toConstr (MyRecord 5 "Hello" "World"))

This will give me the following:

["field1","field2","field3"]

How can I do the same thing for the values in a record, like this:

["5","Hello","World"]

I'm asking because I'm using Aeson to take simple JSON like this:

{
  "field1":5,
  "field2":"Hello",
  "field3":"World"
}

And generate Haskell code like this:

field1 :: Int
field1 = 5

field2 :: String
field2 = "Hello"

field3 :: String
field3 = "World"

How can I generically unwrap all the values in a given record in the same way I can unwrap the field names of my records?

解决方案

Your first question can be answered. If you're happy to convert all the values of a datatype to strings, then you can indeed produce such a list with a generic programming library such as e.g. generics-sop:

{-# LANGUAGE DeriveGeneric, FlexibleContexts #-}

import qualified GHC.Generics as G
import Generics.SOP

data MyRecord = MyRecord
  { field1 :: Int
  , field2 :: String
  , field3 :: String
  } deriving (Show, Eq, G.Generic)

instance Generic MyRecord

stringValues ::
     (Generic a, All2 Show (Code a))
  => a -> [String]
stringValues a =
  hcollapse (hcmap (Proxy :: Proxy Show) (\ (I x) -> K (show x)) (from a))

test :: [String]
test = stringValues (MyRecord 5 "Hello" "world")

Now in GHCi:

GHCi> test
["5","\"Hello\"","\"world\""]

However, if your goal is to maintain the original types, then this is more difficult, as the result type will have to be a heterogeneous list (which in fact generics-sop uses internally, before it is converted back into a normal list using hcollapse).

It's not entirely clear to me what you really want to achieve. It's quite possible that there's a much more straight-forward solution.

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