'Newtonsoft.Json.Linq.JObject'不包含'status'错误的定义

问题描述 投票:3回答:1
responseString = {"status":"success","data":{"text":"Say Hello to\n\nAfrica’sfil’SI\nAnonymous\n\nSocial Network\nFrom Great Ho\nMitt\/6.6om","progress":100}}

dynamic jsonDe = JsonConvert.DeserializeObject(responseString);
            Debug.WriteLine(jsonDe.data);

我究竟做错了什么?

c# windows-phone-8
1个回答
0
投票

最后,这对我有用:

public class  responseData 
{
  public string status { get; set; }
  public dataClass data { get; set; }

}

public class dataClass 
{
public string text { get; set; }
public string progress { get; set; }
}
string myjsonstring = responseString;
            responseData jsonDe = JsonConvert.DeserializeObject<responseData>(myjsonstring);
            Debug.WriteLine(jsonDe.data.progress); // Outputs 100
            Debug.WriteLine(jsonDe.status); // Outputs success
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