Question

我有一个简单的对象，我的API返回

{
"Code":0,
"Message":"String",
"Data":"Can Be Anything"
}

数据可以是字符串或任何其他对象，如人物，猫，列表等......

我将响应映射到c＃，将Data作为Object，假设我将此对象称为“Response”

我想知道在反序列化时是否可行我告知数据类型的类型，并让我们理解并将数据转换为我传递的类型。

会像

public static Response ToWrapper<T>(this string json)
{
    return JsonConvert.DeserializeObject<Response>(json, new Converter<T>());
}

在这个例子中我会说Data是一个名为Person的虚拟类，其中包含道具字符串Name，字符串Email和int Age，所以响应将是

string json = "
{
   "Code":1,
   "Message":"OK",
   "Data": {"Name": "Patrick", "Email": "aa@aaa.com", "Age" : 25}
}"

所以在我的代码的某些方面我会

var ResponseWrapper = json.ToWrapper<Person>();

string personName = ResponseWrapper.Data.Name;  
string personEmail = ResponseWrapper.Data.Email;
int personAge = ResponseWrapper.Data.Age; //<----- NO CAST NEEDED

转换器会将对象数据转换为Person Data !!!

转换器我尝试了一些没有成功的代码！我尝试了很多代码，我可以从这里得到的结果How to implement custom JsonConverter in JSON.NET to deserialize a List of base class objects?

public class PersonConverter<T> : JsonCreationConverter<T>
    {
        public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
        {
            throw new NotImplementedException();
        }

        protected override T Create(Type objectType, JObject jObject)
        {
            T obj = Activator.CreateInstance<T>();

            if (FieldExists("Data", jObject))
            {
                return obj;
            }
            else
                return obj;
        }

        private bool FieldExists(string fieldName, JObject jObject)
        {
            return jObject[fieldName] != null;
        }
    }

    public abstract class JsonCreationConverter<T> : JsonConverter
    {
        /// <summary>
        /// Create an instance of objectType, based properties in the JSON object
        /// </summary>
        /// <param name="objectType">type of object expected</param>
        /// <param name="jObject">
        /// contents of JSON object that will be deserialized
        /// </param>
        /// <returns></returns>
        protected abstract T Create(Type objectType, JObject jObject);

        public override bool CanConvert(Type objectType)
        {
            return true; // typeof(T).IsAssignableFrom(objectType);
        }

        public override bool CanWrite
        {
            get { return false; }
        }

        public override object ReadJson(JsonReader reader,
                                        Type objectType,
                                         object existingValue,
                                         JsonSerializer serializer)
        {
            // Load JObject from stream
            JObject jObject = JObject.Load(reader);

            // Create target object based on JObject
            T target = Create(objectType, jObject);

            // Populate the object properties
            serializer.Populate(jObject.CreateReader(), target);

            return target;
        }
    }

Answer 1

我认为最好的前进方法是将Code和Message的反序列化卸载到JObject，然后使用JsonConvert反序列化数据对象。

https://dotnetfiddle.net/bP2Ew6

using System;
using Newtonsoft.Json;
using Newtonsoft.Json.Linq;

public class Program
{
    public static void Main()
    {
        string json = @"
            {
               ""Code"":1,
               ""Message"":""OK"",
               ""Data"": {""Name"": ""Patrick"", ""Email"": ""aa@aaa.com"", ""Age"" : 25}
            }";

        Console.WriteLine(json);

        var response=json.ToWrapper<Person>();

        Console.WriteLine("Name: "+response.Data.Name);
        Console.WriteLine("Email: "+response.Data.Email);
        Console.WriteLine("Age: "+response.Data.Age);
    }
}

public class Person{
    public string Name {get;set;}
    public string Email {get;set;}
    public int Age {get;set;}
}

public class Response<T>{
    public int Code {get;set;}
    public string Message {get;set;}
    public T Data {get;set;}
}

public static class ResponseExtensions {
    public static Response<T> ToWrapper<T>(this string json){

        var o=JObject.Parse(json);
        var data=JsonConvert.DeserializeObject<T>(o["Data"].ToString());

        return new Response<T>{
            Code=(int)o["Code"],
            Message=(string)o["Message"],
            Data=data
        };
    }
}

C＃Newtonsoft Json转换器将Generic反向串行化为具体的混凝土

1 个答案: