如何使用System.Text.json将json字符串或流反序列化到Dictionary

Question

我得到一个json字符串作为流，并试图将其反序列化为字典

带.NET 6的Program.cs：

using System;
using System.Text;
using System.Text.Json;
using System.Text.Json.Serialization;


Console.WriteLine("Converting Json...");
var result = await DeserializeJson();
Console.WriteLine($"result: {result}");


async Task<Dictionary<string, string>> DeserializeJson()
{
    var jsonText = "{\"number\": 709, \"message\": \"My message here\",\"bool\": true}";
    var stream = new MemoryStream(Encoding.UTF8.GetBytes(jsonText));

    var options = new JsonSerializerOptions 
    { 
         PropertyNameCaseInsensitive = true,
         NumberHandling = JsonNumberHandling.WriteAsString 
    };
    var fullResponse = await JsonSerializer.DeserializeAsync<Dictionary<string, string>>(stream, options);

    return fullResponse;
}

导致的主要错误：

---> System.InvalidOperationException: Cannot get the value of a token type 'Number' as a string.

Answer 1

json不仅仅包含字符串。作为一个快速（但不是性能）修复，您可以尝试反序列化到字典

async Task<Dictionary<string, string>> DeserializeJson()
{
    var jsonText = "{\"number\": 709, \"message\": \"My message here\",\"bool\": true}";
    var stream = new MemoryStream(Encoding.UTF8.GetBytes(jsonText));

    var options = new JsonSerializerOptions 
    { 
        PropertyNameCaseInsensitive = true,
        NumberHandling = JsonNumberHandling.WriteAsString ,
        
    };
    var fullResponse = await JsonSerializer.DeserializeAsync<Dictionary<string, Object>>(stream, options);

    return fullResponse?.ToDictionary(pair => pair.Key, pair => pair.ToString());
}

或手动分析文档：

async Task<Dictionary<string, string>> DeserializeJson()
{
    var jsonText = "{\"number\": 709, \"message\": \"My message here\",\"bool\": true}";
    var stream = new MemoryStream(Encoding.UTF8.GetBytes(jsonText));
  
    var jsonDocument = JsonDocument.Parse(stream);
    var dictionary = jsonDocument.RootElement
        .EnumerateObject()
        .ToDictionary(property => property.Name, property => property.Value.ToString());
    return dictionary;
}