Overwrite Json property name in c#

Question

I have a class with following fields. Those properties are used to serialize as json object when it needs to call a external rest API method.

public class Customer
    {
        [JsonProperty(PropertyName = "email")]
        public string Email { get; set; }

        [JsonProperty(PropertyName = "prop[listId]")]
        public string Test{ get; set; }

        // there are lot of properties 
    }

In the property name Test ,external API service call required some thing like following json filed name format.

prop[7]

In my case this 7 can be changed according the environment like test,dev and prod.So what I'm looking for a way to move that listId value into app.config .

I have tried to do it as follow but it is not allowed to do that.for the listIdValue if assign the constant value it will work.

     private string listIdValue = ConfigurationManager.AppSettings["ListIdValue"];

     [JsonProperty(PropertyName = "prop["+listIdValue +"]")]
     public string Test{ get; set; }

Answer 1

You'll have to override DefaultContractResolver and implement your own mechanism to provide the PropertyName (in JSON). I will provide a full example code to show deserialization and serialization with a runtime generated PropertyName. Currently, it modifies the Test field to Test5 (in all models). You should implement your own mechanism (using an attribute, a reserved name, a table or whatever.

class Program
{
    static void Main(string[] args)
    {
        var customer = new Customer() {Email = "asd@asd.com", Test = "asdasd"};
        var a = Serialize(customer, false);
        var b = Serialize(customer, true);
        Console.WriteLine(a);
        Console.WriteLine(b);

        var desA = Deserialize<Customer>(a, false);
        var desB = Deserialize<Customer>(b, true);

        Console.WriteLine("TestA: {0}", desA.Test);
        Console.WriteLine("TestB: {0}", desB.Test);

    }

    static string Serialize(object obj, bool newNames)
    {
        JsonSerializerSettings settings = new JsonSerializerSettings();
        settings.Formatting = Formatting.Indented;
        if (newNames)
        {
            settings.ContractResolver = new CustomNamesContractResolver();
        }

        return JsonConvert.SerializeObject(obj, settings);
    }
    static T Deserialize<T>(string text, bool newNames)
    {
        JsonSerializerSettings settings = new JsonSerializerSettings();
        settings.Formatting = Formatting.Indented;
        if (newNames)
        {
            settings.ContractResolver = new CustomNamesContractResolver();
        }

        return JsonConvert.DeserializeObject<T>(text, settings);
    }
}
class CustomNamesContractResolver : DefaultContractResolver
{
    protected override IList<JsonProperty> CreateProperties(System.Type type, MemberSerialization memberSerialization)
    {
        // Let the base class create all the JsonProperties 
        // using the short names
        IList<JsonProperty> list = base.CreateProperties(type, memberSerialization);

        // Now inspect each property and replace the 
        // short name with the real property name
        foreach (JsonProperty prop in list)
        {
            if (prop.UnderlyingName == "Test") //change this to your implementation!
                prop.PropertyName = "Test" + 5;

        }

        return list;
    }
}

public class Customer
{
    [JsonProperty(PropertyName = "email")]
    public string Email { get; set; }

    public string Test { get; set; }

}

Output:

{
  "email": "asd@asd.com",
  "Test": "asdasd"
}
{
  "email": "asd@asd.com",
  "Test5": "asdasd"
}
TestA: asdasd
TestB: asdasd

As you see, when we use Serialize(..., false) - the field's name is Test and when we use Serialize(..., true) - the field's name is Test5, as expected. This also works for deserialization.

I have used this answer as insperation for my answer: https://stackoverflow.com/a/20639697/773879