我有以下代码:

var user = (Dictionary<string, object>)serializer.DeserializeObject(responsecontent);

responsecontent中的输入是JSON,但它没有正确地反序列化为对象。我应该如何正确地反序列化它?


当前回答

我假设你没有使用Json。净(Newtonsoft。Json NuGet包)。如果是这样的话,你应该试试。

它具有以下特点:

LINQ到JSON JsonSerializer用于快速将.NET对象转换为JSON并再转换回来 Json。NET可以选择生成格式良好、缩进的JSON以供调试或显示 可以将JsonIgnore和JsonProperty等属性添加到类中,以自定义类的序列化方式 能够将JSON转换为XML 支持多种平台:.NET、Silverlight和Compact Framework

请看下面的例子。在本例中,JsonConvert类用于将对象与JSON进行转换。它有两个用于此目的的静态方法。它们是SerializeObject(Object obj)和DeserializeObject<T>(String json):

using Newtonsoft.Json;

Product product = new Product();
product.Name = "Apple";
product.Expiry = new DateTime(2008, 12, 28);
product.Price = 3.99M;
product.Sizes = new string[] { "Small", "Medium", "Large" };

string json = JsonConvert.SerializeObject(product);
//{
//  "Name": "Apple",
//  "Expiry": "2008-12-28T00:00:00",
//  "Price": 3.99,
//  "Sizes": [
//    "Small",
//    "Medium",
//    "Large"
//  ]
//}

Product deserializedProduct = JsonConvert.DeserializeObject<Product>(json);

其他回答

 using (var ms = new MemoryStream(Encoding.Unicode.GetBytes(user)))
 {
    // Deserialization from JSON  
    DataContractJsonSerializer deserializer = new DataContractJsonSerializer(typeof(UserListing))
    DataContractJsonSerializer(typeof(UserListing));
    UserListing response = (UserListing)deserializer.ReadObject(ms);

 }

 public class UserListing
 {
    public List<UserList> users { get; set; }      
 }

 public class UserList
 {
    public string FirstName { get; set; }       
    public string LastName { get; set; } 
 }

试试下面的代码:

HttpWebRequest request = (HttpWebRequest)WebRequest.Create("URL");
JArray array = new JArray();
using (var twitpicResponse = (HttpWebResponse)request.GetResponse())
using (var reader = new StreamReader(twitpicResponse.GetResponseStream()))
{
    JavaScriptSerializer js = new JavaScriptSerializer();
    var objText = reader.ReadToEnd();

    JObject joResponse = JObject.Parse(objText);
    JObject result = (JObject)joResponse["result"];
    array = (JArray)result["Detail"];
    string statu = array[0]["dlrStat"].ToString();
}

这里有一些不使用第三方库的选项:

// For that you will need to add reference to System.Runtime.Serialization
var jsonReader = JsonReaderWriterFactory.CreateJsonReader(Encoding.UTF8.GetBytes(@"{ ""Name"": ""Jon Smith"", ""Address"": { ""City"": ""New York"", ""State"": ""NY"" }, ""Age"": 42 }"), new System.Xml.XmlDictionaryReaderQuotas());

// For that you will need to add reference to System.Xml and System.Xml.Linq
var root = XElement.Load(jsonReader);
Console.WriteLine(root.XPathSelectElement("//Name").Value);
Console.WriteLine(root.XPathSelectElement("//Address/State").Value);

// For that you will need to add reference to System.Web.Helpers
dynamic json = System.Web.Helpers.Json.Decode(@"{ ""Name"": ""Jon Smith"", ""Address"": { ""City"": ""New York"", ""State"": ""NY"" }, ""Age"": 42 }");
Console.WriteLine(json.Name);
Console.WriteLine(json.Address.State);

有关System.Web.Helpers.Json的更多信息,请参阅链接。

更新:现在最简单的上网方式。helper是使用NuGet包的。


如果你不关心早期的windows版本,你可以使用windows . data . json命名空间的类:

// minimum supported version: Win 8
JsonObject root = Windows.Data.Json.JsonValue.Parse(jsonString).GetObject();
Console.WriteLine(root["Name"].GetString());
Console.WriteLine(root["Address"].GetObject()["State"].GetString());

正如这里的回答-反序列化JSON到c#动态对象?

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;

或者使用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;

另一个本地解决方案是JavaScriptSerializer,它不需要任何第三方库,只需要对System.Web.Extensions的引用。这不是一个新功能,而是一个自3.5以来非常不为人知的内置功能。

using System.Web.Script.Serialization;

..

JavaScriptSerializer serializer = new JavaScriptSerializer();
objectString = serializer.Serialize(new MyObject());

和背部

MyObject o = serializer.Deserialize<MyObject>(objectString)