Detect if deserialized object is missing a field with the JsonConvert class in Json.NET

The Json.Net serializer has a MissingMemberHandling setting which you can set to Error. (The default is Ignore.) This will cause the serializer to throw a JsonSerializationException during deserialization whenever it encounters a JSON property for which there is no corresponding property in the target class.

static void Main(string[] args)
{
    try
    {
        JsonSerializerSettings settings = new JsonSerializerSettings();
        settings.MissingMemberHandling = MissingMemberHandling.Error;

        var goodObj = JsonConvert.DeserializeObject<MyJsonObjView>(correctData, settings);
        System.Console.Out.WriteLine(goodObj.MyJsonInt.ToString());

        var badObj = JsonConvert.DeserializeObject<MyJsonObjView>(wrongData, settings);
        System.Console.Out.WriteLine(badObj.MyJsonInt.ToString());
    }
    catch (Exception ex)
    {
        Console.WriteLine(ex.GetType().Name + ": " + ex.Message);
    }
}

Result:

42
JsonSerializationException: Could not find member 'SomeOtherProperty' on object
of type 'MyJsonObjView'. Path 'SomeOtherProperty', line 3, position 33.

See: MissingMemberHandling setting.

Just add [JsonProperty(Required = Required.Always)] to the required properties and it'll throw exception if the property is not there while deserializing.

[JsonProperty(Required = Required.Always)]
 public int MyJsonInt { get; set; }

Put the following attribute on required properties:

[DataMember(IsRequired = true)]

If the member is not present, it will throw a Newtonsoft.Json.JsonSerializationException.

As Brian suggested below, you will also need this attribute on your class:

[DataContract]