.NET核心的答案

你可以使用内置的System.Text.Json而不是第三方的Json.NET。为了促进重用，json文件读取功能属于自己的类，应该是泛型的，而不是硬编码为某种类型(Item)。下面是一个完整的例子:

using System;
using System.IO;
using System.Text.Json;
using System.Threading.Tasks;

namespace Project
{
    class Program
    {
        static async Task Main()
        {
            Item item = await JsonFileReader.ReadAsync<Item>(@"C:\myFile.json");
        }
    }

    public static class JsonFileReader
    {
        public static async Task<T> ReadAsync<T>(string filePath)
        {
            using FileStream stream = File.OpenRead(filePath);
            return await JsonSerializer.DeserializeAsync<T>(stream);
        }
    }

    public class Item
    {
        public int millis;
        public string stamp;
        public DateTime datetime;
        public string light;
        public float temp;
        public float vcc;
    }
}

或者，如果你更喜欢简单的/同步的:

class Program
{
    static void Main()
    {
        Item item = JsonFileReader.Read<Item>(@"C:\myFile.json");
    }
}

public static class JsonFileReader
{
    public static T Read<T>(string filePath)
    {
        string text = File.ReadAllText(filePath);
        return JsonSerializer.Deserialize<T>(text);
    }
}

有一个更简单的方法从文件或从Web获取JSON: Json.Net.Curl

安装包Json.Net.Curl

// get JObject from local file system 
var json = Json.Net.Curl.Get(@"data\JObjectUnitTest1.json");
var json = await Json.Net.Curl.GetAsync(@"data\JObjectUnitTest1.json")


// get JObject from Server  
var json = await Json.Net.Curl.GetAsync("http://myserver.com/data.json");

GitHub项目 Nuget

基于@ l.b.。的解决方案，(类型为对象而不是匿名)VB代码是

Dim oJson As Object = JsonConvert.DeserializeObject(File.ReadAllText(MyFilePath))

我应该提到，对于构造不需要类型的HTTP调用内容，这是快速且有用的。使用对象而不是匿名意味着你可以在Visual Studio环境中保持Option Strict On -我讨厌关闭它。

使用开源库Cinchoo ETL，解析非常大的JSON文件是迭代的，使用简单

1. 动态方法:—不需要POCO类

        string json = @"
[
  {
    ""millis"": ""1000"",
    ""stamp"": ""1273010254"",
    ""datetime"": ""2010/5/4 21:57:34"",
    ""light"": ""333"",
    ""temp"": ""78.32"",
    ""vcc"": ""3.54""
  },
  {
    ""millis"": ""2000"",
    ""stamp"": ""1273010254"",
    ""datetime"": ""2010/5/4 21:57:34"",
    ""light"": ""333"",
    ""temp"": ""78.32"",
    ""vcc"": ""3.54""
  }
] 
";
        
        using (var r = ChoJSONReader.LoadText(json))
        {
            foreach (var rec in r)
                Console.WriteLine(rec.Dump());
        }

样本提琴:https://dotnetfiddle.net/mo1qvw

2. POCO:受压迫的

定义匹配json属性的POCO类

public class Item
{
    public int Millis { get; set; }
    public string Stamp { get; set; }
    public DateTime Datetime { get; set; }
    public string Light { get; set; }
    public float Temp { get; set; }
    public float Vcc { get; set; }
}

然后使用解析器加载JSON，如下所示

        string json = @"
[
  {
    ""millis"": ""1000"",
    ""stamp"": ""1273010254"",
    ""datetime"": ""2010/5/4 21:57:34"",
    ""light"": ""333"",
    ""temp"": ""78.32"",
    ""vcc"": ""3.54""
  },
  {
    ""millis"": ""2000"",
    ""stamp"": ""1273010254"",
    ""datetime"": ""2010/5/4 21:57:34"",
    ""light"": ""333"",
    ""temp"": ""78.32"",
    ""vcc"": ""3.54""
  }
] 
";
        
        using (var r = ChoJSONReader<Item>.LoadText(json))
        {
            foreach (var rec in r)
                Console.WriteLine(ChoUtility.Dump(rec));
        }

样本提琴:https://dotnetfiddle.net/fRWu0w

免责声明:我是这个库的作者。

Doing this yourself is an awful idea. Use Json.NET. It has already solved the problem better than most programmers could if they were given months on end to work on it. As for your specific needs, parsing into arrays and such, check the documentation, particularly on JsonTextReader. Basically, Json.NET handles JSON arrays natively and will parse them into strings, ints, or whatever the type happens to be without prompting from you. Here is a direct link to the basic code usages for both the reader and the writer, so you can have that open in a spare window while you're learning to work with this.

这是最好的:这次偷懒，使用一个库，这样你就永远解决了这个常见的问题。

这段代码可以帮助你:

string _filePath = Path.GetDirectoryName(System.AppDomain.CurrentDomain.BaseDirectory);

JObject data = JObject.Parse(_filePath );