如何在没有 JSON.NET 库的情况下解析 JSON? [英] How to parse JSON without JSON.NET library?

问题描述

我正在尝试在 Visual Studio 2011 上为 Windows 8 构建 Metro 应用程序.当我尝试这样做时，我在如何在没有 JSON.NET 库的情况下解析 JSON 时遇到了一些问题(它还不支持 Metro 应用程序)).

I'm trying to build a Metro application for Windows 8 on Visual Studio 2011. and while I'm trying to do that, I'm having some issues on how to parse JSON without JSON.NET library (It doesn't support the metro applications yet).

无论如何，我想解析这个:

Anyway, I want to parse this:

{
   "name":"Prince Charming",
   "artist":"Metallica",
   "genre":"Rock and Metal",
   "album":"Reload",
   "album_image":"http://up203.siz.co.il/up2/u2zzzw4mjayz.png",
   "link":"http://f2h.co.il/7779182246886"
}

推荐答案

您可以使用 System.Json 命名空间.您需要添加对 System.Runtime.Serialization 程序集

You can use the classes found in the System.Json Namespace which were added in .NET 4.5. You need to add a reference to the System.Runtime.Serialization assembly

JsonValue.Parse() 方法 解析 JSON 文本并返回 JsonValue:

The JsonValue.Parse() Method parses JSON text and returns a JsonValue:

JsonValue value = JsonValue.Parse(@"{ ""name"":""Prince Charming"", ...");

如果您传递带有 JSON 对象的字符串，您应该能够将该值转换为 JsonObject:

If you pass a string with a JSON object, you should be able to cast the value to a JsonObject:

using System.Json;


JsonObject result = value as JsonObject;

Console.WriteLine("Name .... {0}", (string)result["name"]);
Console.WriteLine("Artist .. {0}", (string)result["artist"]);
Console.WriteLine("Genre ... {0}", (string)result["genre"]);
Console.WriteLine("Album ... {0}", (string)result["album"]);

这些类与系统中的类非常相似.Xml.Linq 命名空间.

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