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
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