With System.Text.Json, is there a way to EXCLUDE properties from a BASE class from serialization?

Question

For example, I have a class that derives from ObservableValidator and my resulting serialization then includes HasErrors. I want to only include my derived class's properties and don't have access to modify the base class. Is there a way to do this?

public class Derived: ObservableValidator {

    [Required]
    [JsonPropertyName("name")] 
    [ObservableProperty]
    private string? _name;

}

serialized json:

 {
  "name": "Test",
  "HasErrors": false
}

How can I make name the only property written to my json?

NewtonSoft has what I want with this decorator, but haven't found an equivalent in System.Text.Json.

[JsonObject(MemberSerialization.OptIn)]
public class File

Could be related to similar issue, I have in addition to this one in which the MVVM generator doesn't propagate the [JsonIgnore]. Maybe need to refactor with non-observable record type class. Is there any other way of ignoring a property during JSON serialization instead of using [JsonIgnore] decorator?

"$type": "Derived", will only show up if you enable support for polymorphism -- and it's not enabled by default. So how are you enabling it? Does the ObservableValidator type do that, by applying [JsonDerivedType(typeof(Derived))]? Or are you doing it through options somehow? Can you share a minimal reproducible example? — dbc
– dbc, Commented Feb 26 at 23:51
Thank you dbc, I actually had just realized that's where $type came from and fixed that part. Derived also implemented a small interface (that I can edit) which is where it was from. As I don't need to deserialize, it wasn't needed. — azulBonnet
– azulBonnet, Commented Feb 26 at 23:54
You could use the typeInfo modifier JsonExtensions.IgnorePropertiesDeclaredBy<ObservableValidator>() from this answer to Serializing/Deserializing an object that is derived from Java.Lang.Object throws exception (using System.Text.Json) to skip serialization of all properties declared by ObservableValidator. Does that answer your question? — dbc
– dbc, Commented Feb 27 at 0:13
Did Serializing/Deserializing an object that is derived from Java.Lang.Object throws exception (using System.Text.Json) answer your question, or do you still need some additional help? — dbc
– dbc, Commented Feb 27 at 17:52
Thank you, that does work for me. If you want to submit an answer, I will accept it. — azulBonnet
– azulBonnet, Commented Feb 27 at 19:36

Andrew KeepCoding · Accepted Answer · 2025-02-28 05:36:39Z

3

There's an open issue about this. Voting up there might speed up the fix.

As a work around, you can use a custom type info resolver:

public sealed class IgnoreHasErrorsTypeResolver : DefaultJsonTypeInfoResolver
{
    public override JsonTypeInfo GetTypeInfo(Type type, JsonSerializerOptions options)
    {
        JsonTypeInfo jsonTypeInfo = base.GetTypeInfo(type, options);

        if (jsonTypeInfo
            .Properties
            .FirstOrDefault(property => property.Name is nameof(ObservableValidator.HasErrors)) is JsonPropertyInfo hasErrorsJsonPropertyInfo)
        {
            jsonTypeInfo.Properties.Remove(hasErrorsJsonPropertyInfo);
        }

        return jsonTypeInfo;
    }
}

and use it like:

JsonSerializerOptions options = new()
{
    TypeInfoResolver = new IgnoreHasErrorsTypeResolver(),
};

string jsonText = JsonSerializer.Serialize(someObject, options);

edited Feb 28 at 5:36

answered Feb 28 at 5:24

Andrew KeepCoding

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1 Comment

Latency Jun 20 at 1:05

This worked for me! Thank you... Now I can take my DependancyProperty class and serialize it.

chandanbsd · Accepted Answer · 2025-02-27 05:35:49Z

I would create a model/dto and map the object from the class in question to the dto.

For example:

public class Derived: ObservableValidator {
    [Required]
    [ObservableProperty]
    public string? _name;
}

public class DerivedModel {
    [JsonPropertyName("name")] 
    public string? _name;
}


// Usage:

var originalObject = new Derived()
originalObject.name = "Hello World!";

var derivedObject = new DerivedModel();
originalObject.name = originalObject.name;

Now apply the serialization on the derivedObject.

Do note that the above code is a basic example and can be significantly improved upon by

using property instead of fields
Using https://automapper.org/ to simplify conversions from one user-defined type to another.
even using records for the model instead of class

Let me know if you'd like a more elaborate example with the above implementations.

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