How to change Json serialization of only one field using Json.net?

Question

Say, for example, I'm trying to convert an object with 10 fields to Json, however I need to change the serialization process of 1 of these fields. At this point, I will have to manually write each property as follows:

 public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        writer.WriteStartObject();

        writer.WritePropertyName("Field1");
        serializer.Serialize(writer, value.Field1);

        writer.WritePropertyName("Field2");
        serializer.Serialize(writer, value.Field2);

        writer.WritePropertyName("Field3");
        serializer.Serialize(writer, value.Field3);

        writer.WritePropertyName("Field4");
        serializer.Serialize(writer, Convert.ToInt32(value.Field4)); //Modifying one field here

        //Six more times

        writer.WriteEndObject();
    }

This is not good code, and it is really annoying to write. Is there a way to get Json.net to automatically serialize everything but one property? Or perhaps generate a JObject automatically and change this?

Answer 1

You can try decorating the property that you want to change manually with JsonConverterAttributeand pass the appropriate type JsonConverter.

, OP:

public class IntegerConverter : JsonConverter
{
  public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
  {
    serializer.Serialize(writer, Convert.ToInt32(value));
  }

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

  public override bool CanConvert(Type objectType)
  {
    return objectType == typeof(string);
  }
}

class TestJson
{
  public string Field1 { get; set; }
  public string Field2 { get; set; }
  public string Field3 { get; set; }

  [JsonConverter(typeof(IntegerConverter))]
  public string Field4 { get; set; }        
}

, JsonConvert:

var test = new TestJson {Field1 = "1", Field2 = "2", Field3 = "3", Field4 = "4"};

var jsonString = JsonConvert.SerializeObject(test);

Answer 2

( ), . , this.is :

public class MyClass
{
    public string Value4 {get; set;}
}

, 4 int, :

public class MyClass
{
    [JsonIgnore()]
    public string Value4 {get; set;}

    public int Value4AsInt
    {
        return Convert.ToInt32(Value4);
    }
}

Answer 3

You can use System.Reflection, however it is slow, but you do not need to change the class

public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
{
    writer.WriteStartObject();

    Type vType = value.GetType();
    MemberInfo[] properties = vType.GetProperties(BindingFlags.Public
                                        | BindingFlags.Instance);

    foreach (PropertyInfo property in properties)
    {
        object serValue = null;
        if (property.Name == "Field4")
        {
            serValue = Convert.ToInt32(property.GetValue(value, null));
        }
        else
        {
            serValue = property.GetValue(value, null);
        }
        writer.WritePropertyName(property.Name);
        serializer.Serialize(writer, serValue);
    }

    writer.WriteEndObject();
}