CODEWITHSUNDEEP
asp.net-mvc
Kevin himch
Kevin himch

Reputation: 297

Grab json response from api

Please check the controller bellow. I have to process the Json response inside this controller. My goal is to grab "success" value true or false. How can i access that value from following json response? I already tried to use JsSerialializer but no luck. Please check pictures for better understanding. Thanks in advance.

View Model:

public class GoogleRecaptcha
{
    public bool success { get; set; }
    public DateTime challenge_ts { get; set; }
    public string hostname { get; set; }
}

pic1 pic2

public ActionResult DownloadProcess()
{
    using (BlexzWebDbEntities db = new BlexzWebDbEntities())
    {
        //Validate Google recaptcha below
        var Response = Request["g-recaptcha-response"];
        string SecretKey = "6Lghu-MJoniMPXVf";

        //var client = new WebClient();

        var client = new RestClient("https://www.google.com/recaptcha/api/siteverify");
        var request = new RestRequest(Method.POST);
        request.Method = Method.POST;
        request.Parameters.Clear();
        request.AddParameter("secret", SecretKey);
        request.AddParameter("response", Response);
        var ResultFromGoogle = client.Execute(request).Content;

        JavaScriptSerializer jsSerializer = new JavaScriptSerializer();
        var r = jsSerializer.DeserializeObject(ResultFromGoogle);


        return View();
    }

}

Upvotes: 0

Views: 72

Answers (1)

CodeFuller
CodeFuller

Reputation: 31312

There are multiple ways to do what you want:

1. Using JavaScriptSerializer with strongly typed model:

public class ResultFromGoogle
{
    public bool Success { get; set; }

    public DateTime Challenge_ts { get; set; }

    public string Hostname { get; set; }
}

ResultFromGoogle results = jsSerializer.Deserialize<ResultFromGoogle>(ResultFromGoogle);
var successValue = results.Success;

2. Using JavaScriptSerializer without strongly typed model:

var props = (Dictionary<string, object>)jsSerializer.DeserializeObject(ResultFromGoogle);
var successValue = props["success"];

3. Using JSON.Net with strongly typed model:

var results = JsonConvert.DeserializeObject<ResultFromGoogle>();
var successValue = results.Success;

4. Using JSON.Net without strongly typed model:

var successValue = JObject.Parse(ResultFromGoogle).GetValue("success");

Overall, I suggest using strongly-typed model, because it provides conversion to required type (e.g. bool for success property and DateTime for challenge_ts). It's also more convenient if you need to access other fields of the response.

Json.Net in its turn is much more flexible than JavaScriptSerializer. You need to install Json.Net NuGet to use options #3 and #4.

Upvotes: 1

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