rwvens/JsonKnownTypes

Simple way to serialize and deserialize polymorphic types for Json.NET

JsonKnownTypes .Net Standard

Help to serialize and deserialize polymorphic types. Add distractor to json data

• Documentation
• License

Requirements

• NET Standard 2.0 compatible project
• Json.NET by Newtonsoft

Documentation

Getting started

There is simple way to use just add one attribute to base class or interface

  [JsonConverter(typeof(JsonKnownTypesConverter<BaseClass>))]
  public class BaseClass
  {
    public string Summary { get; set; }
  }
  
  public class ChildClass : BaseClass
  {
    public string Detailed { get; set; }
  }

Serialization and Deserialization:

  var entityJson = JsonConvert.SerializeObject(entity);
  var obj = DeserializeObject<BaseClass>(entityJson)

Json representation:

{ "Summary":"someValue", "$type":"BaseClass" }
{ "Summary":"someValue", "Detailed":"someValue", "$type":"ChildClass" }

Using with Interfaces or Abstract classes

Also you can use it similar with interfaces or abstract classes

Interface

  [JsonConverter(typeof(JsonKnownTypesConverter<BaseClass>))]
  public interface IInterface  { ... }
 
  public class ChildClass : IInterface  { ... }

Json representation:

{ ... "$type":"ChildClass" }

Abstract class

  [JsonConverter(typeof(JsonKnownTypesConverter<BaseClass>))]
  public abstract class AbstractClass  { ... }
 
  public class ChildClass : AbstractClass  { ... }

Json representation:

{ ... "$type":"ChildClass" }

JsonKnownType

If you need to add custom discriminator just use JsonKnowType attribute.
By default for discriminattor property using "$type" name, if you need to change it use JsonDiscriminator attribute.

  [JsonConverter(typeof(JsonKnownTypesConverter<BaseClass>))]
  [JsonDiscriminator(Name = "myType")] //add custom discriminator name
  [JsonKnownType(typeof(BaseClass1Heir))] //could be deleted if you didn't turn off AutoJsonKnownType
  [JsonKnownType(typeof(BaseClass2Heir), "myDiscriminator")]
  public class BaseClass { ... }
  
  public class BaseAbstractClass1Heir : BaseClass  { ... }
  
  public class BaseAbstractClass2Heir : BaseClass  { ... }

Json representation:

{ ... , "myType":"BaseClass" }

{ ... , "myType":"BaseClass1Heir" }

{ ... , "myType":"myDiscriminator" }

JsonKnownThisType

Add discriminator for type which is used with it

  [JsonConverter(typeof(JsonKnownTypesConverter<BaseClass>))]
  [JsonKnownThisType("do_you_know_that")]
  public class BaseClass { ... }
  
  [JsonKnownThisType("html_is_programming_language")]
  public class BaseAbstractClass1Heir : BaseClass  { ... }
  
  [JsonKnownThisType("just_joke=)")]
  public class BaseAbstractClass2Heir : BaseClass  { ... }

Json representation:

{ ... , "$type":"do_you_know_that" }

{ ... , "$type":"html_is_programming_language" }

{ ... , "$type":"just_joke=)" }

Configuration

For change default discriminator settings use:

  JsonKnownTypesSettingsManager.DefaultDiscriminatorSettings = new JsonDiscriminatorSettings
  {
    DiscriminatorFieldName = "name",
    UseClassNameAsDiscriminator = false
  };

DiscriminatorFieldName change default "$type" name to yours

If UseClassNameAsDiscriminator is false you should to add JsonKnownType or JsonKnownThisType attribute for each relative class manualy or it throw an Exception

Use manualy

  public class BaseClass { ... }
  public class BaseAbstractClass1Heir : BaseClass  { ... }
  public class BaseAbstractClass2Heir : BaseClass  { ... }

  var converter = new JsonKnownTypesConverter<BaseClass>()
  
  var entityJson = JsonConvert.SerializeObject(entity, converter);
  var obj = DeserializeObject<BaseClass>(entityJson, converter)

Need to add converter to method just if you don't use JsonConverter attribute

License

Authored by: Dmitry Kaznacheev (dmitry-bym)

This project is under MIT license. You can obtain the license copy here.

This work using work of James Newton-King, author of Json.NET. https://www.newtonsoft.com/json