Elm：如何 Json 使用单个 TypeConstructor 解码联合类型？

Question

如以下代码所示，解码 UserAlias 很容易，但是当我尝试解码 UserType 时，即用 D.map2 UserType 替换 D.map2 UserAlias 编译器大声喊叫。如何修复此编译器错误？

import Json.Decode as D
import Html exposing (..)
import Result as R

type  UserType = UserType {name:String, age:Int}

type alias UserAlias = {name:String, age:Int}


userDecoder = D.map2 UserAlias
                (D.field "name" D.string)
                (D.field "age" D.int)


decodeUser json = D.decodeString userDecoder json 


json = """
  { "name": "Bob", "age": 40 }
"""


main = div [] [(text << toString << decodeUser) json]

以上代码运行良好。现在将 D.map2 UserAlias 替换为 D.map2 UserType

编译器哭了

Detected errors in 1 module.
==================================== ERRORS ====================================



-- TYPE MISMATCH ---------------------------------------------------------------

The 2nd argument to function `map2` is causing a mismatch.

13|               D.map2 UserType
14|>                (D.field "name" D.string)
15|                 (D.field "age" D.int)

Function `map2` is expecting the 2nd argument to be:

    D.Decoder { age : Int, name : String }

But it is:

    D.Decoder String

Hint: I always figure out the type of arguments from left to right. If an
argument is acceptable when I check it, I assume it is "correct" in subsequent
checks. So the problem may actually be in how previous arguments interact with
the 2nd.



-- TYPE MISMATCH ---------------------------------------------------------------

The 1st argument to function `map2` is causing a mismatch.

13|>              D.map2 UserType
14|                 (D.field "name" D.string)
15|                 (D.field "age" D.int)

Function `map2` is expecting the 1st argument to be:

    { age : Int, name : String } -> b -> UserType

But it is:

    { age : Int, name : String } -> UserType

Hint: It looks like a function needs 1 more argument.

如何解决这个错误，请帮忙！

基本上我不再希望使用类型别名，而只希望使用用户 UserType，这样我就可以隐藏内部记录结构，并在不破坏 public API.

的情况下重构它

谢谢。

Answer 1

UserType 构造函数采用单个参数 UserAlias，因此我们可以简单地在现有解码器中使用 Json.Decode.map 来获取 UserAlias 值并构造解码器对于 UserType 这样的：

userDecoder : D.Decoder UserType
userDecoder = D.map2 UserAlias
                (D.field "name" D.string)
                (D.field "age" D.int)
                |> D.map UserType

Answer 2

比 Chad 的回答更直接的方法是在 map2 函数中执行此操作：

userDecoder : D.Decoder UserType
userDecoder = D.map2 (\name age -> UserType <| UserAlias name age)
                (D.field "name" D.string)
                (D.field "age" D.int)

这将避免第二个 map 中的模式匹配。不过，就性能而言，它可能不会给你带来太多好处！

如果你觉得更简洁一点，这也可以写成无点风格：

userDecoder = D.map2 (((<<) UserType ) << UserAlias)
                (D.field "name" D.string)
                (D.field "age" D.int)

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