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How Can I Dynamically Deserialize JSON in C# Using Json.Net?

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Release: 2025-02-01 00:41:11
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How Can I Dynamically Deserialize JSON in C# Using Json.Net?

Use json.net dynamic derivatives json

When processing JSON data, sometimes it is helpful to serialize the data into a dynamic object. This allows you to access attributes without the need to explicitly specify their types.

Use dynamic back serialization

JSON.NET provides the function of using Dynamic keywords to sequence JSON's back series into dynamic objects:

This code assumes that the JSON string JSON contains attributes called MESSAGE.

dynamic jsonResponse = JsonConvert.DeserializeObject(json);
Console.WriteLine(jsonResponse.message);
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<示> Example

Consider the following JSON data:

You can sequence this JSON's back series into dynamic objects as follows:

{
  "number": 1000,
  "str": "string",
  "array": [1,2,3,4,5,6]
}
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<多> More information

dynamic d = JObject.Parse("{number:1000, str:'string', array: [1,2,3,4,5,6]}");

Console.WriteLine(d.number); // 输出:1000
Console.WriteLine(d.str);    // 输出:string
Console.WriteLine(d.array.Count);  // 输出:6
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For more information about using json.net for dynamic JSON back -sequentialization, see JSON.NET's Linq to JSON document.

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