multiple JsonProperty Name assigned to single property

Tricking custom JsonConverter worked for me. Thanks @khaled4vokalz, @Khanh TO

public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        object instance = objectType.GetConstructor(Type.EmptyTypes).Invoke(null);
        PropertyInfo[] props = objectType.GetProperties();

        JObject jo = JObject.Load(reader);
        foreach (JProperty jp in jo.Properties())
        {
            if (string.Equals(jp.Name, "name1", StringComparison.OrdinalIgnoreCase) || string.Equals(jp.Name, "name2", StringComparison.OrdinalIgnoreCase))
            {
                PropertyInfo prop = props.FirstOrDefault(pi =>
                pi.CanWrite && string.Equals(pi.Name, "CodeModel", StringComparison.OrdinalIgnoreCase));

                if (prop != null)
                    prop.SetValue(instance, jp.Value.ToObject(prop.PropertyType, serializer));
            }
        }

        return instance;
    }

I had the same use case, though in Java.

Resource that helped https://www.baeldung.com/json-multiple-fields-single-java-field

We can use a

@JsonProperty("main_label_to_serialize_and_deserialize")
@JsonAlias("Alternate_label_if_found_in_json_will_be_deserialized")

In your use case you could do

@JsonProperty("name1")
@JsonAlias("name2")

A simple solution which does not require a converter: just add a second, private property to your class, mark it with [JsonProperty("name2")], and have it set the first property:

public class Specifications
{
    [JsonProperty("name1")]
    public string CodeModel { get; set; }

    [JsonProperty("name2")]
    private string CodeModel2 { set { CodeModel = value; } }
}

Fiddle: https://dotnetfiddle.net/z3KJj5