如何调用 JsonConvert.DeserializeObject 并禁用通过 [JsonConverter] 应用于基本类型的 JsonConverter?

How to call JsonConvert.DeserializeObject and disable a JsonConverter applied to a base type via [JsonConverter]?(如何调用 JsonConvert.DeserializeObject 并禁用通过 [JsonConverter] 应用于基本类型的 JsonConverter?)

问题描述

澄清问题:

我已经为基类型覆盖了 JsonConverter(通过将 [JsonConverter(typeof(TConverter))] 应用于超类)，但是当直接反序列化子类型时，我想使用 STANDARD 序列化(即没有自定义转换器)用于反序列化我的派生对象.如何在反序列化方法中指定 STANDARD 序列化，就好像我没有覆盖 JsonConverter 一样?

I have overridden the JsonConverter for a base type (by applying [JsonConverter(typeof(TConverter))] to the superclass), but when deserializing the sub-type directly I want to use STANDARD serialization (i.e. no custom converter) for deserializing my derived object. How do I specify STANDARD serialization for use in the deserialize method, as if I had NOT overridden the JsonConverter?

我正在使用弹性搜索，无法使用我的自定义 JsonConverter 实现调用 JsonConvert.DeserializeObject，并且必须依赖 Elastic 的属性才能使用我的转换器.

I am using elastic search and can't call JsonConvert.DeserializeObject with my custom implementation of JsonConverter, and have to rely on the attribute for Elastic to use my converter.

但是，使用这个转换器作为属性似乎也会影响所有子类，但我只是希望它们使用标准转换器，这样我就不必为许多实现中的每一个都实现 JsonConverter.

However, using this converter as attribute seems to affect all sub classes as well, but I just want them to use the standard converter, so that I don't have to implement JsonConverter for each of many implementations.

这是我想要的类/逻辑:

This is my classes/logic as I would like it to look:

    [Route("test")]
    [HttpPost]
    public HttpResponseMessage Test([FromBody] JToken json)
    {
        var res = json.ToObject<Product>(); // I want an object of ProductImpl type here
        return Request.CreateResponse(res); 
    }

    [JsonConverter(typeof(JsonProductConverted))]
    public abstract class Product
    {
    }

    public class ProductImpl : Product
    {
    }

    public class JsonProductConverted : JsonConverter
    {
        public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
        {
            throw new NotImplementedException();
        }

        public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
        {
            JObject json = JObject.Load(reader);
            //var type = GetTypeFromId((int) json["typeId"]); // Construct type from field in 
            var type = typeof(ProductImpl);
            // var res = JsonConvert.DeserializeObject(json.ToString(), type, DEFAULT_JSONCONVERTER_HERE);
            var res = DeserializeToObjectWithStandardJsonConverter(json, type);
            return res;
        }

        public override bool CanConvert(Type objectType)
        {
            return false;
        }
    }

如果我不提供默认的 JsonConverter 或类似的，它将只使用 JsonProductConverted 转换器，这会创建一个无限循环.

If I don't supply the default JsonConverter, or similar it will just use the JsonProductConverted converter, which creates an infinite loop.

推荐答案

由于您已将 [JsonConverter(typeof(JsonProductConverted))] 直接添加到您的 Product 类型中，您可以向 ProductImpl 添加一个虚拟转换器，该转换器从 false" rel="nofollow noreferrer">CanRead 和 CanWrite:

Since you have added [JsonConverter(typeof(JsonProductConverted))] directly to your Product type, you could add a dummy converter to ProductImpl that returns false from CanRead and CanWrite:

[JsonConverter(typeof(NoConverter))]
public class ProductImpl : Product
{
}

public class NoConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return false;
    }

    public override bool CanRead { get { return false; } }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }

    public override bool CanWrite { get { return false; } }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

这会覆盖基类的转换器，然后回退到读取和写入的默认序列化

This overrides the base class's converter and then falls back on default serialization for both reading and writing

示例 .Net fiddle.

另一种选择是使用 serializer.Populate().这避免了为对象本身调用转换器:

public class JsonProductConverted : JsonTypeInferringConverterBase
{
    protected override Type InferType(Type objectType, JObject json)
    {
        //var type = GetTypeFromId((int) json["typeId"]); // Construct type from field in 
        return typeof(ProductImpl);
    }

    public override bool CanConvert(Type objectType)
    {
        return false;
    }
}

public abstract class JsonTypeInferringConverterBase : JsonConverter
{
    public override bool CanWrite { get { return false; } }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }

    protected abstract Type InferType(Type objectType, JObject json);

    protected virtual object CreateObject(Type actualType, JsonSerializer serializer, JObject json)
    {
        var contract = (JsonObjectContract)serializer.ContractResolver.ResolveContract(actualType);
        return contract.DefaultCreator();
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        if (reader.TokenType == JsonToken.Null)
            return null;
        var json = JObject.Load(reader);

        var actualType = InferType(objectType, json);

        // Construct object (or reuse existingValue if compatible)
        if (existingValue == null || !actualType.IsAssignableFrom(existingValue.GetType()))
        {
            existingValue = CreateObject(actualType, serializer, json);
        }

        // Populate object.
        using (var subReader = json.CreateReader())
        {
            serializer.Populate(subReader, existingValue);
        }

        return existingValue;
    }
}