c# - 解析 json 对象

Question

我无法理解如何使用 Visual .NET 将 JSON 字符串解析为 c# 对象。任务很简单，但我还是迷路了……我得到了这个字符串：

{"single_token":"842269070","username":"example123","version":1.1}

这是我尝试消毒的代码：

namespace _SampleProject
{
    public partial class Downloader : Form
    {
        public Downloader(string url, bool showTags = false)
        {
            InitializeComponent();
            WebClient client = new WebClient();
            string jsonURL = "http://localhost/jev";   
            source = client.DownloadString(jsonURL);
            richTextBox1.Text = source;
            JavaScriptSerializer parser = new JavaScriptSerializer();
            parser.Deserialize<???>(source);
        }

我不知道在“<”和“>”之间放什么，从我在网上阅读的内容来看，我必须为它创建一个新类..？另外，我如何获得输出？一个例子会很有帮助！

Answer 1

创建一个可以将您的 JSON 反序列化为的新类，例如：

public class UserInfo
{
    public string single_token { get; set; }
    public string username { get; set; }
    public string version { get; set; }
}

public partial class Downloader : Form
{
    public Downloader(string url, bool showTags = false)
    {
        InitializeComponent();
        WebClient client = new WebClient();
        string jsonURL = "http://localhost/jev";
        source = client.DownloadString(jsonURL);
        richTextBox1.Text = source;
        JavaScriptSerializer parser = new JavaScriptSerializer();
        var info = parser.Deserialize<UserInfo>(source);

        // use deserialized info object
    }
}

Answer 2

如果您使用的是 .NET 4 - 使用动态数据类型。

http://msdn.microsoft.com/en-us/library/dd264736.aspx

string json = "{ single_token:'842269070', username: 'example123', version:1.1}";

     JavaScriptSerializer jss = new JavaScriptSerializer();

     dynamic obj = jss.Deserialize<dynamic>(json);

     Response.Write(obj["single_token"]);
     Response.Write(obj["username"]);
     Response.Write(obj["version"]); 

Answer 3

是的，您需要一个具有与您的 JSON 匹配的属性的新类。

MyNewClass result = parser.Deserialize<MyNewClass>(source);

Answer 4

通常的方法是创建一个类（或一组类，用于更复杂的 JSON 字符串）来描述您要反序列化的对象并将其用作通用参数。

另一种选择是将 JSON 反序列化为Dictionary<string, object>：

parser.Deserialize<Dictionary<string, object>>(source);

这样，您可以访问数据，但除非您必须这样做，否则我不建议您这样做。

Answer 5

您需要一个与您获得的 JSON 匹配的类，它将返回该类的一个新对象，其中填充了值。

Answer 6

以下是代码..

ServicePointManager.ServerCertificateValidationCallback = new System.Net.Security.RemoteCertificateValidationCallback(AcceptAllCertifications);

        request = WebRequest.Create("https://myipaddress/api/admin/configuration/v1/conference/1/");

        request.Credentials = new NetworkCredential("admin", "admin123");
        // Create POST data and convert it to a byte array.
        request.Method = "GET";          

                // Set the ContentType property of the WebRequest.
        request.ContentType = "application/json; charset=utf-8";          


        WebResponse response = request.GetResponse();
        // Display the status.
        Console.WriteLine(((HttpWebResponse)response).StatusDescription);
        // Get the stream containing content returned by the server.
        dataStream = response.GetResponseStream();
        // Open the stream using a StreamReader for easy access.
        StreamReader reader = new StreamReader(dataStream);
        // Read the content.
        string responseFromServer = reader.ReadToEnd();
        JavaScriptSerializer js = new JavaScriptSerializer();
        var obj = js.Deserialize<dynamic>(responseFromServer);
        Label1.Text = obj["name"];
        // Display the content.
        Console.WriteLine(responseFromServer);
        // Clean up the streams.
        reader.Close();
        dataStream.Close();
        response.Close();

Answer 7

dynamic data = JObject.Parse(jsString);
var value= data["value"];