是否有方法将JSON内容反序列化为c#动态类型?为了使用DataContractJsonSerializer,最好跳过创建一堆类。


当前回答

我使用http://json2csharp.com/来获取表示JSON对象的类。

输入:

{
   "name":"John",
   "age":31,
   "city":"New York",
   "Childs":[
      {
         "name":"Jim",
         "age":11
      },
      {
         "name":"Tim",
         "age":9
      }
   ]
}

输出:

public class Child
{
    public string name { get; set; }
    public int age { get; set; }
}

public class Person
{
    public string name { get; set; }
    public int age { get; set; }
    public string city { get; set; }
    public List<Child> Childs { get; set; }
}

之后我使用Newtonsoft。Json填充类:

using Newtonsoft.Json;

namespace GitRepositoryCreator.Common
{
    class JObjects
    {
        public static string Get(object p_object)
        {
            return JsonConvert.SerializeObject(p_object);
        }
        internal static T Get<T>(string p_object)
        {
            return JsonConvert.DeserializeObject<T>(p_object);
        }
    }
}

你可以这样调用它:

Person jsonClass = JObjects.Get<Person>(stringJson);

string stringJson = JObjects.Get(jsonClass);

PS:

如果你的JSON变量名不是一个有效的c#名称(名称以$开头),你可以这样修复:

public class Exception
{
   [JsonProperty(PropertyName = "$id")]
   public string id { get; set; }
   public object innerException { get; set; }
   public string message { get; set; }
   public string typeName { get; set; }
   public string typeKey { get; set; }
   public int errorCode { get; set; }
   public int eventId { get; set; }
}

其他回答

JsonFx可以将JSON内容反序列化为动态对象。

序列化动态类型(.NET 4.0的默认值):

var reader = new JsonReader(); var writer = new JsonWriter();

string input = @"{ ""foo"": true, ""array"": [ 42, false, ""Hello!"", null ] }";
dynamic output = reader.Read(input);
Console.WriteLine(output.array[0]); // 42
string json = writer.Write(output);
Console.WriteLine(json); // {"foo":true,"array":[42,false,"Hello!",null]}

你可以在Newtonsoft.Json的帮助下实现这一点。从NuGet安装它,然后:

using Newtonsoft.Json;

dynamic results = JsonConvert.DeserializeObject<dynamic>(YOUR_JSON);

最简单的方法是:

只需包含这个DLL文件。

像这样使用代码:

dynamic json = new JDynamic("{a:'abc'}");
// json.a is a string "abc"

dynamic json = new JDynamic("{a:3.1416}");
// json.a is 3.1416m

dynamic json = new JDynamic("{a:1}");
// json.a is

dynamic json = new JDynamic("[1,2,3]");
/json.Length/json.Count is 3
// And you can use json[0]/ json[2] to get the elements

dynamic json = new JDynamic("{a:[1,2,3]}");
//json.a.Length /json.a.Count is 3.
// And you can use  json.a[0]/ json.a[2] to get the elements

dynamic json = new JDynamic("[{b:1},{c:1}]");
// json.Length/json.Count is 2.
// And you can use the  json[0].b/json[1].c to get the num.

使用Json非常简单。NET:

dynamic stuff = JsonConvert.DeserializeObject("{ 'Name': 'Jon Smith', 'Address': { 'City': 'New York', 'State': 'NY' }, 'Age': 42 }");

string name = stuff.Name;
string address = stuff.Address.City;

同样使用Newtonsoft.Json.Linq:

dynamic stuff = JObject.Parse("{ 'Name': 'Jon Smith', 'Address': { 'City': 'New York', 'State': 'NY' }, 'Age': 42 }");

string name = stuff.Name;
string address = stuff.Address.City;

文档:使用动态查询JSON

在JSON中反序列化。NET可以使用包含在该库中的JObject类来实现动态。我的JSON字符串表示这些类:

public class Foo {
   public int Age {get;set;}
   public Bar Bar {get;set;}
}

public class Bar {
   public DateTime BDay {get;set;}
}

现在我们在不引用上述类的情况下反序列化字符串:

var dyn = JsonConvert.DeserializeObject<JObject>(jsonAsFooString);

JProperty propAge = dyn.Properties().FirstOrDefault(i=>i.Name == "Age");
if(propAge != null) {
    int age = int.Parse(propAge.Value.ToString());
    Console.WriteLine("age=" + age);
}

//or as a one-liner:
int myage = int.Parse(dyn.Properties().First(i=>i.Name == "Age").Value.ToString());

或者如果你想深入一点:

var propBar = dyn.Properties().FirstOrDefault(i=>i.Name == "Bar");
if(propBar != null) {
    JObject o = (JObject)propBar.First();
    var propBDay = o.Properties().FirstOrDefault (i => i.Name=="BDay");
    if(propBDay != null) {
        DateTime bday = DateTime.Parse(propBDay.Value.ToString());
        Console.WriteLine("birthday=" + bday.ToString("MM/dd/yyyy"));
    }
}

//or as a one-liner:
DateTime mybday = DateTime.Parse(((JObject)dyn.Properties().First(i=>i.Name == "Bar").First()).Properties().First(i=>i.Name == "BDay").Value.ToString());

完整的示例请参见文章。