How to prevent object deserialization with Json.Net?

Question

JSON

I need to get an object from a Json file with the following structure

{
"start" : 1,
"finish": 2,
"template" : 3
}

Here is my C# class:

public sealed class ORequest
    {
        public ORequest() { }

        [JsonProperty("start")]
        public long? start { get; set; }

       [JsonProperty("finish")]
        public long? finish { get; set; }

        [JsonProperty("template")]
       public int? template { get; set; }
    }

I create an object of class ORequest :

ORequest outRequest = JsonConvert.DeserializeObject<ORequest>(message, new JsonSerializerSettings { 
    NullValueHandling = NullValueHandling.Ignore,  
    MissingMemberHandling = MissingMemberHandling.Ignore
 });

But the deserialization is performed even if I put the following code in the message:

{
"sta" : 1,
"fini": 2,
"temp" : 3
}

As a result, an outRequest object will be created where all fields will be null.
How to make JsonConvertDeserilize return null for invalid fields in Json file

Answer 1

Try Required :

[JsonProperty(PropertyName = "start", Required = Required.Always)]
public long? start { get; set; }

Or check the value of the key field (or fields), if the value is null , consider that the deserialization did not produce results ( outRequest = null ).
A key field is a field that is always present and cannot be null .
If there is nothing to catch on, then you can go through everyone:

if (!outRequest.start.HasValue && !outRequest.finish.HasValue && !outRequest.template.HasValue)
{
  outRequest = null;
}

In theory, if key fields use a type other than Nullable (for example, long instead of long? ), or together with the Required.Always attribute , then an exception should be thrown during deserialization if the required fields in the data are not found. An exception can be caught ( try-catch ) and thus understand that the incoming data is incorrect.
You can also make a custom converter - JsonConverter , but this can be a difficult decision, in terms of "energy costs" :-)

[JsonConverter(typeof(ORequestConverter))]
public sealed class ORequest
{
  [JsonProperty("start")]
  public long? start { get; set; }

  [JsonProperty("finish")]
  public long? finish { get; set; }

  [JsonProperty("template")]
  public int? template { get; set; }
}

public class ORequestConverter : JsonConverter
{

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

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

    var target = (ORequest)Activator.CreateInstance(objectType);

    serializer.Populate(reader, target);

    if (!target.start.HasValue && !target.finish.HasValue && !target.template.HasValue)
    {
      // необходимые поля не найдены или имеют значение null
      // возвращаем null
      return null;
    }

    return target;
  }

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

  public override bool CanConvert(Type objectType)
  {
    throw new NotImplementedException();
  }

}

Answer 2

JsonIgnore attribute?