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

Posted 2021-05-04

responseString = {"status":"success","data":{"text":"Say Hello to

Africa’sfil’SI
Anonymous

Social Network
From Great Ho
Mitt/6.6om","progress":100}}

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

我究竟做错了什么？

答案

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