Question

我正在读取我的xamarin.forms android应用程序中的webservice的响应，以下是包含状态（0-error，1-OK）消息的响应＆amp; info（info包含来自datatable的数据行）

{
"status": 1,
"msg" : "OK",
"info": {
"UCode": "1",
"UName": "Admin",
"UPass": "pass"
}
}

我能够阅读status＆amp; msg。

如何将数据从节点信息转换为类User_Info的可观察集合？

这是我的代码

            try
            {
                using (var client = new HttpClient())
                {
                    var url = GSVar.hostname + GSVar.user_check;
                    var content = new FormUrlEncodedContent(new[]
                    {
                        new KeyValuePair<string,string>("uname",T1.Text),
                        new KeyValuePair<string, string>("upass",T2.Text)
                    });

                    var resp = await client.PostAsync(new Uri(url), content);
                    //var resp = await client.GetAsync(new Uri(url));
                    if (resp.IsSuccessStatusCode)
                    {
                        var result = JsonConvert.DeserializeObject<Json_Respnce>(resp.Content.ReadAsStringAsync().Result);
                        if (result.status == 0)
                            General.GSErr(result.msg);
                        else
                        {
                            //User_Info user_info = JsonConvert.DeserializeObject<User_Info>(result.UserInfo);
                            //await DisplayAlert("OK", result.UserInfo.ToString(), "OK");
                        }
                    }
                    else
                        General.GSErr("Nothing retrieved from server.");
                }
            }
            catch { throw; }

列出类

class Json_Respnce
{
    [JsonProperty(PropertyName ="status")]
    public int status { get; set; }

    [JsonProperty(PropertyName = "msg")]
    public string msg { get; set; }

    //[JsonProperty(PropertyName = "info")]
    //public string UserInfo { get; set; }
}

class User_Info
{
    [JsonProperty(PropertyName = "UCode")]
    public string UCode { get; set; }

    [JsonProperty(PropertyName = "UName")]
    public string UName { get; set; }

    [JsonProperty(PropertyName = "UPass")]
    public string UPass { get; set; }
}

Answer 1

创建所需的模型类。您可以使用json2csharp。只需将JSON字符串粘贴到那里，然后单击Generate

public class Info
{
    public string UCode { get; set; }
    public string UName { get; set; }
    public string UPass { get; set; }
}

public class Response
{
    public int status { get; set; }
    public string msg { get; set; }
    public Info info { get; set; }
}

然后你可以将你的JSON字符串反序列化为：

string jsonString = await resp.Content.ReadAsStringAsync ();
Response response = JsonConvert.DeserializeObject<Response> (jsonString));

if (response.status == 1) {
    Info info = response.info
}

阅读C＃Xamarin.Forms中的JSON响应

1 个答案: