Derived type's properties missing in JSON response from ASP.NET Core API

Question

The JSON response from my ASP.NET Core 3.1 API controller is missing properties. This happens when a property uses a derived type; any properties defined in the derived type but not in the base/interface will not be serialized to JSON. It seems there is some lack of support for polymorphism in the response, as if serialization is based on a property's defined type instead of its runtime type. How can I change this behavior to ensure that all public properties are included in the JSON response?

Example:

My .NET Core Web API Controller returns this object that has a property with an interface type.

    // controller returns this object
    public class Result
    {
        public IResultProperty ResultProperty { get; set; }   // property uses an interface type
    }

    public interface IResultProperty
    { }

Here is a derived type that defines a new public property named Value.

    public class StringResultProperty : IResultProperty
    {
        public string Value { get; set; }
    }

If I return the derived type from my controller like this:

    return new MainResult {
        ResultProperty = new StringResultProperty { Value = "Hi there!" }
    };

then the actual response includes an empty object (the Value property is missing):

I want the response to be:

    {
        "ResultProperty": { "Value": "Hi there!" }
    }

Answer 1

I ended up creating a custom JsonConverter (System.Text.Json.Serialization namespace) which forces JsonSerializer to serialize to the object's runtime type. See the Solution section below. It's lengthy but it works well and does not require me to sacrifice object oriented principles in my API's design. (If you need something quicker and can use Newtonsoft then check out the top voted answer instead.)

Some background: Microsoft has a System.Text.Json serialization guide with a section titled Serialize properties of derived classes with good information relevant to my question. In particular it explains why properties of derived types are not serialized:

This behavior is intended to help prevent accidental exposure of data in a derived runtime-created type.

If that is not a concern for you then the behavior can be overridden in the call to JsonSerializer.Serialize by either explicitly specifying the derived type or by specifying object, for example:

    // by specifying the derived type
    jsonString = JsonSerializer.Serialize(objToSerialize, objToSerialize.GetType(), serializeOptions);
    
    // or specifying 'object' works too
    jsonString = JsonSerializer.Serialize<object>(objToSerialize, serializeOptions);

To accomplish this with ASP.NET Core you need to hook into the serialization process. I did this with a custom JsonConverter that calls JsonSerializer.Serialize one of the ways shown above. I also implemented support for deserialization which, while not explicitly asked for in the original question, is almost always needed anyway. (Oddly, supporting only serialization and not deserialization proved to be tricky anyway.)

Solution

I created a base class, DerivedTypeJsonConverter, which contains all of the serialization & deserialization logic. For each of your base types, you would create a corresponding converter class for it that derives from DerivedTypeJsonConverter. This is explained in the numbered directions below.

This solution follows the "type name handling" convention from Json.NET which introduces support for polymorphism to JSON. It works by including an additional $type property in the derived type's JSON (ex: "$type":"StringResultProperty") that tells the converter what the object's true type is. (One difference: in Json.NET, $type's value is a fully qualified type + assembly name, whereas my $type is a custom string which helps future-proof against namespace/assembly/class name changes.) API callers are expected to include $type properties in their JSON requests for derived types. The serialization logic solves my original problem by ensuring that all of the object's public properties are serialized, and for consistency the $type property is also serialized.

Directions:

1) Copy the DerivedTypeJsonConverter class below into your project.

    using System;
    using System.Collections.Generic;
    using System.Dynamic;
    using System.IO;
    using System.Linq;
    using System.Reflection;
    using System.Text;
    using System.Text.Json;
    using System.Text.Json.Serialization;

    public abstract class DerivedTypeJsonConverter<TBase> : JsonConverter<TBase>
    {
        protected abstract string TypeToName(Type type);
    
        protected abstract Type NameToType(string typeName);
    

        private const string TypePropertyName = "$type";
    

        public override bool CanConvert(Type objectType)
        {
            return typeof(TBase) == objectType;
        }
    
    
        public override TBase Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options)
        {
            // get the $type value by parsing the JSON string into a JsonDocument
            JsonDocument jsonDocument = JsonDocument.ParseValue(ref reader);
            jsonDocument.RootElement.TryGetProperty(TypePropertyName, out JsonElement typeNameElement);
            string typeName = (typeNameElement.ValueKind == JsonValueKind.String) ? typeNameElement.GetString() : null;
            if (string.IsNullOrWhiteSpace(typeName)) throw new InvalidOperationException($"Missing or invalid value for {TypePropertyName} (base type {typeof(TBase).FullName}).");
    
            // get the JSON text that was read by the JsonDocument
            string json;
            using (var stream = new MemoryStream())
            using (var writer = new Utf8JsonWriter(stream, new JsonWriterOptions { Encoder = options.Encoder })) {
                jsonDocument.WriteTo(writer);
                writer.Flush();
                json = Encoding.UTF8.GetString(stream.ToArray());
            }
    
            // deserialize the JSON to the type specified by $type
            try {
                return (TBase)JsonSerializer.Deserialize(json, NameToType(typeName), options);
            }
            catch (Exception ex) {
                throw new InvalidOperationException("Invalid JSON in request.", ex);
            }
        }
    

        public override void Write(Utf8JsonWriter writer, TBase value, JsonSerializerOptions options)
        {
            // create an ExpandoObject from the value to serialize so we can dynamically add a $type property to it
            ExpandoObject expando = ToExpandoObject(value);
            expando.TryAdd(TypePropertyName, TypeToName(value.GetType()));
    
            // serialize the expando
            JsonSerializer.Serialize(writer, expando, options);
        }
    

        private static ExpandoObject ToExpandoObject(object obj)
        {
            var expando = new ExpandoObject();
            if (obj != null) {
                // copy all public properties
                foreach (PropertyInfo property in obj.GetType().GetProperties(BindingFlags.Public | BindingFlags.Instance).Where(p => p.CanRead)) {
                    expando.TryAdd(property.Name, property.GetValue(obj));
                }
            }
    
            return expando;
        }
    }

2) For each of your base types, create a class that derives from DerivedTypeJsonConverter. Implement the 2 abstract methods which are for mapping $type strings to actual types. Here is an example for my IResultProperty interface that you can follow.

    public class ResultPropertyJsonConverter : DerivedTypeJsonConverter<IResultProperty>
    {
        protected override Type NameToType(string typeName)
        {
            return typeName switch
            {
                // map string values to types
                nameof(StringResultProperty) => typeof(StringResultProperty)

                // TODO: Create a case for each derived type
            };
        }
    
        protected override string TypeToName(Type type)
        {
            // map types to string values
            if (type == typeof(StringResultProperty)) return nameof(StringResultProperty);

            // TODO: Create a condition for each derived type
        }
    }

3) Register the converters in Startup.cs.

    services.AddControllers()
        .AddJsonOptions(options => {
            options.JsonSerializerOptions.Converters.Add(new ResultPropertyJsonConverter());

            // TODO: Add each converter
        });

4) In requests to the API, objects of derived types will need to include a $type property. Example JSON: { "Value":"Hi!", "$type":"StringResultProperty" }

Full gist here