通过Json.Net反序列化F#Map

时间:2014-02-13 07:20:00

标签: json serialization f# json.net

我有一个简单的问题:是否可以从解析F#Map类型?因为当我尝试它时(使用F# Map<string, string>),它很容易序列化并且它看起来如何,但是当我尝试反序列化时它会抛出异常。

Newtonsoft.Json.JsonSerializationException: Unable to find a default constructor to use for type Microsoft.FSharp.Collections.FSharpMap`2[System.Int32,System.String]. Path '1', line 2, position 7.
  at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.CreateNewDictionary (Newtonsoft.Json.JsonReader reader, Newtonsoft.Json.Serialization.JsonDictionaryContract contract, System.Boolean& createdFromNonDefaultConstructor) [0x00000] in <filename unknown>:0 
  at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.CreateObject (Newtonsoft.Json.JsonReader reader, System.Type objectType, Newtonsoft.Json.Serialization.JsonContract contract, Newtonsoft.Json.Serialization.JsonProperty member, Newtonsoft.Json.Serialization.JsonContainerContract containerContract, Newtonsoft.Json.Serialization.JsonProperty containerMember, System.Object existingValue) [0x00000] in <filename unknown>:0 
  at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.CreateValueInternal (Newtonsoft.Json.JsonReader reader, System.Type objectType, Newtonsoft.Json.Serialization.JsonContract contract, Newtonsoft.Json.Serialization.JsonProperty member, Newtonsoft.Json.Serialization.JsonContainerContract containerContract, Newtonsoft.Json.Serialization.JsonProperty containerMember, System.Object existingValue) [0x00000] in <filename unknown>:0 
  at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.Deserialize (Newtonsoft.Json.JsonReader reader, System.Type objectType, Boolean checkAdditionalContent) [0x00000] in <filename unknown>:0

它是经典的反序列化:

Map.ofList [ ("1", "one"); ("2", "two"); ("3", "three") ]

生成的JSON看起来像C#字典

{
  "1": "one",
  "2": "two",
  "3": "three"
}

它没有设置序列化(只有缩进)。那么可以序列化这个,还是有一些可行的解决方法?

感谢您的回答

4 个答案:

答案 0 :(得分:3)

您可以制作自己的转换器来执行此操作。它有很多反思和构建适当的泛型类型,但它可以完成。

您首先反序列化为Dictionary<Key, Val>,然后通过反射手动创建并填充List<Tuple<Key, Val>>(因为Map构造函数需要Tuples,而不是KeyValuePairs) ,然后最终将其传递给Map构造函数。

不确定是否有更简单的方法,但这就是我提出的方法:

open System
open System.Collections
open System.Collections.Generic
open Newtonsoft.Json

let mapConverter = {
  new JsonConverter() with

    override x.CanConvert(t:Type) =
      t.IsGenericType && t.GetGenericTypeDefinition() = typedefof<Map<_, _>>

    override x.WriteJson(writer, value, serializer) =
      serializer.Serialize(writer, value)

    override x.ReadJson(reader, t, _, serializer) =
      let genArgs = t.GetGenericArguments()
      let generify (t:Type) = t.MakeGenericType genArgs
      let tupleType = generify typedefof<Tuple<_, _>>
      let listType = typedefof<List<_>>.MakeGenericType tupleType
      let create (t:Type) types = (t.GetConstructor types).Invoke
      let list = create listType [||] [||] :?> IList
      let kvpType = generify typedefof<KeyValuePair<_, _>>
      for kvp in serializer.Deserialize(reader, generify typedefof<Dictionary<_, _>>) :?> IEnumerable do
        let get name = (kvpType.GetProperty name).GetValue(kvp, null)
        list.Add (create tupleType genArgs [|get "Key"; get "Value"|]) |> ignore        
      create (generify typedefof<Map<_, _>>) [|listType|] [|list|]
}

转换后,您只需将其传递到DeserializeObject方法,JsonConvert将在适当的位置使用它。

let str = JsonConvert.SerializeObject (Map<_, _> [333, 1234])
JsonConvert.DeserializeObject<Map<int, int>>(str, mapConverter)

这样做的好处是,如果你有一个很大/很深的记录,你的Map只是一个字段,那么它也可以使用它 - 您不必更改记录结构以仅使用Dictionaries来支持序列化。

答案 1 :(得分:2)

此功能在V6.0.3中成为JSON.Net的一部分。 (2014年4月30日)

但是,如果你因为某些原因而使用早期版本而陷入困境,那么Dax Fohl版本的简化(更有效,反映更少)版本可能是:

type mapConvert<'f,'t when 'f : comparison>() =
    static member readJson (reader:JsonReader, serializer:JsonSerializer) =
        serializer.Deserialize<Dictionary<'f, 't>> (reader)
        |> Seq.map (fun kv -> kv.Key, kv.Value)
        |> Map.ofSeq

let mapConverter = {
  new JsonConverter() with
    override __.CanConvert (t:Type) =
      t.IsGenericType && t.GetGenericTypeDefinition() = typedefof<Map<_, _>>

    override __.WriteJson (writer, value, serializer) =
      serializer.Serialize(writer, value)

    override __.ReadJson (reader, t, _, serializer) =
      let converter = 
        typedefof<mapConvert<_,_>>.MakeGenericType (t.GetGenericArguments())

      let readJson =
        converter.GetMethod("readJson")

      readJson.Invoke(null, [| reader; serializer |])
}

答案 2 :(得分:1)

问题是json.net无法构造Map<int,string>。但是,如果您反序列化为常规.net Dictionary<int,string>,它将起作用,因为json是相同的。

答案 3 :(得分:0)

你不能直接序列化F#的Map,因为它根本没有默认的构造函数(没有参数的构造函数)。

这是F#map的原始文档:(来自http://msdn.microsoft.com/en-us/library/ee353686%28v=vs.110%29.aspx

[<Sealed>]
type Map<[<EqualityConditionalOnAttribute>] 'Key,[<ComparisonConditionalOnAttribute>] [<EqualityConditionalOnAttribute>] 'Value (requires comparison)> =
    class
interface IEnumerable
interface IComparable
interface IEnumerable
interface ICollection
interface IDictionary
new Map : seq<'Key * 'Value> -> Map< 'Key, 'Value>
member this.Add : 'Key * 'Value -> Map<'Key, 'Value>
member this.ContainsKey : 'Key -> bool
member this.Remove : 'Key -> Map<'Key, 'Value>
member this.TryFind : 'Key -> 'Value option
member this.Count :  int
member this.IsEmpty :  bool
member this.Item ('Key) : 'Value
end

如上所示,Map没有默认构造函数,但序列化程序需要一个带默认构造函数的类。

序列化地图的最佳方法是将地图映射为常规的.NET字典,但新字典不具备F#Map的所有优点,尤其是F#Map的不变性。