Specifying custom property name when binding objec

Question

I have a .Net Core Web API. It automatically maps models when the model properties match the request body. For example, if you have this class:

public class Package
{
    public string Carrier { get; set; }
    public string TrackingNumber { get; set; }
}

It would correctly bind it to a POST endpoint if the request body is the following JSON:

{
    carrier: "fedex",
    trackingNumber: "123123123"
}

What I need to do is specify a custom property to map. For example, using the same class above, I need to be able to map to JSON if the TrackingNumber comes in as tracking_number.

How do I do that?

Answer 1

For DotnetCore3.1 We can use

public class Package
{

    [JsonProperty("carrier")]
    public string Carrier { get; set; }

    [JsonProperty("trackingNumber")]
    public string TrackingNumber { get; set; }
}

Answer 2

By using custom converter you will be able to achieve what you need.
The following suite of components based on attributes might suit your needs and is quite generic in case you want to expand it.

Base attribute class

Defines the IsMatch which allows you to define if the object property matches the json property.

[AttributeUsage(AttributeTargets.Property | AttributeTargets.Field, AllowMultiple = true)]
public abstract class JsonDeserializationPropertyMatchAttribute : Attribute
{
    protected JsonDeserializationPropertyMatchAttribute() { }

    public abstract bool IsMatch(JProperty jsonProperty);
}

Sample implementation: multi deserialization names

Defines an attribute which allows you to have multiple names associated to a property. The IsMatch implementation simply loops through them and tries to find a match.

[AttributeUsage(AttributeTargets.Property | AttributeTargets.Field, AllowMultiple = true)]
public class JsonDeserializationNameAttribute : JsonDeserializationPropertyNameMatchAttribute
{
    public string[] PropertyNames { get; private set; }

    public JsonDeserializationNameAttribute(params string[] propertyNames)
    {
        this.PropertyNames = propertyNames;
    }

    public override bool IsMatch(JProperty jsonProperty)
    {
        return PropertyNames.Any(x => String.Equals(x, jsonProperty.Name, StringComparison.InvariantCultureIgnoreCase));
    }
}

The Converter in order to bind both attributes to the json deserialization the following converter is required:

public class JsonDeserializationPropertyMatchConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return objectType.IsClass;
    }

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

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        var constructor = objectType.GetConstructor(new Type[0]);
        if (constructor == null)
            throw new JsonSerializationException("A parameterless constructor is expected.");

        var value = constructor.Invoke(null);

        var jsonObject = JObject.Load(reader);
        var jsonObjectProperties = jsonObject.Properties();

        PropertyInfo[] typeProperties = objectType.GetProperties();
        var typePropertyTuples = new List<Tuple<PropertyInfo, Func<JProperty, bool>>>();
        foreach (var property in typeProperties.Where(x => x.CanWrite))
        {
            var attribute = property.GetCustomAttribute<JsonDeserializationPropertyMatchAttribute>(true);
            if (attribute != null)
                typePropertyTuples.Add(new Tuple<PropertyInfo, Func<JProperty, bool>>(property, attribute.IsMatch));
            else
                typePropertyTuples.Add(new Tuple<PropertyInfo, Func<JProperty, bool>>(property, (x) => false));
        }

        foreach (JProperty jsonProperty in jsonObject.Properties())
        {
            var propertyTuple = typePropertyTuples.FirstOrDefault(x => String.Equals(jsonProperty.Name, x.Item1.Name, StringComparison.InvariantCultureIgnoreCase) || x.Item2(jsonProperty));
            if (propertyTuple != null)
                propertyTuple.Item1.SetValue(value, jsonProperty.Value.ToObject(propertyTuple.Item1.PropertyType, serializer));
        }

        return value;
    }

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

Sample using the codes pasted above, and by decorating the class as follow, I managed to get the objects to deserialize properly:

[JsonConverter(typeof(JsonDeserializationPropertyMatchConverter))]
public class Package
{
    public string Carrier { get; set; }

    [JsonDeserializationName("Tracking_Number","anotherName")]
    public string TrackingNumber { get; set; }
}

Output 1

var input = "{ carrier: \"fedex\", trackingNumber: \"123123123\" }";
var output = JsonConvert.DeserializeObject<Package>(input); // output.TrackingNumber is "123123123"

Output 2

var input = "{ carrier: \"fedex\", tracking_Number: \"123123123\" }";
var output = JsonConvert.DeserializeObject<Package>(input); // output.TrackingNumber is "123123123"

Output 3

var input = "{ carrier: \"fedex\", anotherName: \"123123123\" }";
var output = JsonConvert.DeserializeObject<Package>(input); // output.TrackingNumber is "123123123"

Answer 3

In My Case, I don't want to change property name Carrier & TrackingNumber

so I just add new JsonSerializerSettings() on JsonResult response

public JsonResult GetJQXGridData(){
    var Data = .......
    return Json(Data, new JsonSerializerSettings()) //change here
}

Without using JsonSerializerSettings Output

{
    carrier: "fedex",
    trackingNumber: "123123123"
}

With using JsonSerializerSettings Output

{
    Carrier: "fedex",
    TrackingNumber: "123123123"
}

Answer 4

I think that this should work too:

using Microsoft.AspNetCore.Mvc;
public class Package
{
     [BindProperty(Name ="carrier")]
     public string Carrier { get; set; }

     [BindProperty(Name ="trackingNumber")]
     public string TrackingNumber { get; set; }
}

Answer 5

Change your package class and add JsonProperty decoration for each field you wish to map to a different json field.

public class Package
{
    [JsonProperty(PropertyName = "carrier")]
    public string Carrier { get; set; }

    [JsonProperty(PropertyName = "trackingNumber")]
    public string TrackingNumber { get; set; }
}

Answer 6

TejSoft's answer does not work in ASP.NET Core 3.0 Web APIs by default.

Starting in 3.0, the ASP.NET Core Json.NET (Newtonsoft.Json) sub-component is removed from the ASP.NET Core shared framework. It is announced that, "Json.NET will continue to work with ASP.NET Core, but it will not be in the box with the shared framework." The newly added Json Api claimed to be specifically geared for high-performance scenarios.

Use JsonPropertyName attribute to set a custom property name:

using System.Text.Json.Serialization;

public class Package
{
    [JsonPropertyName("carrier")]
    public string Carrier { get; set; }

    [JsonPropertyName("tracking_number")]
    public string TrackingNumber { get; set; }
}

Hope it helps!