Overwrite Json property name in c#

Solution 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 = "[email protected]", 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": "[email protected]",
  "Test": "asdasd"
}
{
  "email": "[email protected]",
  "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

Solution 2:

Define different configuration modes like Debug/Release/QA/Staging

Then add compilation symbols for each one of them. and in your code you do something like:

Following I suppose you defined: QA and STAGING

public class Customer
{

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

    #if QA
        [JsonProperty(PropertyName = "prop[QA_ID]")]
    #elif STAGING
        [JsonProperty(PropertyName = "prop[STAGING_ID]")]
    #endif
    public string Test{ get; set; }

    // there are lot of properties 
}

You can use these configuration for automated deploy too which will save you time.