Question

我有一个数据交换框架，数据将被格式化为JSON，以便使用http从服务器传递给客户端。

这是'基本结构：

sql = 'update Atm set Balance=%s Where ID=%s'%(str(rollingBalance),str(ID))

或：

{
  status: 0,
  data: {name:"balbala",age:12}
  msg: "OK"
}

或者发生错误的事情：

{
  status: 0,
  data: [{name: "balbalba", age:12},{name:"another balbala", age:13}],
  msg: "OK"

}

我们的基本想法是使用{ status: -1,//the error code msg: "Sorry, internal error" }作为错误代码，并在执行更多操作之前检查错误代码，当不是status时，只显示0这是详细信息错误的信息，但在msg字段中我们想要携带任意JSON格式，主要是data和array，我们可以在java中使用object(key-values)实现这一点，但是来到gson我们不知道如何做到这一点，有没有办法用C#做到这一点？

这是我目前实现的，但有一些错误。

c#

这是错误：

//the Response structure Definition.
public class Respose
{
    public int status;
    public object data;
    public string msg;

    public Respose() { }
}

//serialize the HTTP text first time to a Response structure.
Respose resp = JsonConvert.DeserializeObject<Respose>(infoStr);

 //just check the status shutcut the processing when meet with a non-zero err-code
if(resp.status != 0)
{
        Debug.LogWarning(resp.msg);
        yield break;
}
//getting the data as json string so that we could deserialize it again
string levelArrayStr = JsonConvert.SerializeObject(resp.data);

//the real data-load we need.
List<Level> levelsFromServer = JsonConvert.DeserializeObject<List<Level>>(levelArrayStr);

Answer 1

您可以使用两个（或三个）单独的类进行反序列化，也可以使用JObject。

public class Data
{
    public string Name;
    public int Age;
}

const string SINGLE = @"{
  status: 0,
  data: {name: ""balbala"",age:12},
  msg: ""OK""
}";


const string MULTI = @"{
  status: 0,
  data: [{name: ""balbala"",age:12},{name: ""another  balbala"",age:13}],
  msg: ""OK""
}";

const string ERROR = @"{
  status: 1,
  msg: ""Sorry, internal error""
}";

List<Data> loDataList = new List<Data>();
//JObject loJObject = JObject.Parse(ERROR);
//JObject loJObject = JObject.Parse(SINGLE);
JObject loJObject = JObject.Parse(MULTI);
if (loJObject.Value<int>("status") != 0)
{
    Debug.WriteLine(loJObject.Value<string>("msg"));
    return;
}

JToken loDataToken = loJObject["data"];
switch (loDataToken.Type)
{
    case JTokenType.Object:
        loDataList.Add(loJObject["data"].ToObject<Data>());
        break;
    case JTokenType.Array:
        loDataList = loJObject["data"].ToObject<List<Data>>();
        break;
}

foreach (var loData in loDataList)
{
    Debug.WriteLine("Name: {0} Age: {1}", loData.Name, loData.Age);
}

使用Newtonsoft.Json反序列化/序列化任意数据类型

1 个答案: