Add multiple elements in a JSON array for an object in C # using Json.net

Question

Can someone tell me how can I deserialize an object containing multiple attributes?

Based on the scenario below, the code is working fine.

public ActionResult Index()
{
    string json = @"{""name"": ""Person 2"",""email"": ""[e-mail]""}";

    var emp = JsonConvert.DeserializeObject<Person>(json);
    Response.Write(emp.name + emp.email);
    return View();
}

public class Person
{
    public string name { get; set; }
    public string email { get; set; }
}

But what should I do if the array contains several elements, for example

string json = @"{""data"": [{""name"": ""Person 1"",""email"": ""[e-mail]""},{""name"": ""Person 2"",""email"": ""[e-mail]""}]}";

Thanks in advance

The answers below were ideal for the problem I asked, but now I have taken one step forward. Can anyone advise what I need to do if json had an array? add address to?

{
    "data": [
        {
            "name": "Person 1",
            "email": "[e-mail]",
            "address": {
                "address1": "my address 1",
                "address2": "my address 2" 
            } 
        },
        {
            "name": "Person 2",
            "email": "[e-mail]",
            "address": {
                "address1": "my address 1",
                "address2": "my address 2" 
            } 
        } 
    ] 
}

Answer 1

Something like this has worked for me in the past:

JObject o = JObject.Parse(json); // This would be the string you defined above
// The next line should yield a List<> of Person objects...
List<Person> list = JsonConvert.DeserializeObject<List<Person>>(o["data"].ToString());

You might want to decorate your Person object as follows:

[JsonObject(MemberSerialization.OptIn)]
public class Person
{
    [JsonProperty]
    public string name{get;set;}
    [JsonProperty]
    public string email{get;set;}
}

Answer 2

.

var template = new { data = new Person[] { } };
Person[] emps = JsonConvert
    .DeserializeAnonymousType(json, template)
    .data;

Answer 3

deserialize as:

public class JsonData
{
  public List<Person> Data {get;set;}
}