如果问题是将某些对象字段绑定到某些 json 字段：

如果您想指定此字段的路径而不是简单的字段名称，那么此解决方案适合您：

System.Text.Json的解决方案：

我想了很久，还是对标准库做了一个相当简单的解决方案：

public class PathJsonConverter<T> : JsonConverter<T> where T : new()
{
    // Этот метод упрощает создание настроек
    public static T? Deserialize(string json, JsonNamingPolicy? namingPolicy = null) 
    {
        var options = new JsonSerializerOptions();
        options.PropertyNamingPolicy = namingPolicy;
        options.Converters.Add(new PathJsonConverter<T>());
        return JsonSerializer.Deserialize<T>(json, options);
    }

    public override T? Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options)
    {
        var jsonObject = JsonNode.Parse(ref reader);
        
        T? result = new T();

        foreach (var property in typeToConvert.GetProperties())
        {
            JsonNode? currentNode = jsonObject;
            var attr = property.GetCustomAttribute<JsonPropertyNameAttribute>();
            if (attr == null)
                currentNode = currentNode?[options.PropertyNamingPolicy?.ConvertName(property.Name) ?? property.Name];
            else
                foreach (var path in attr.Name.Split('.'))
                    currentNode = currentNode?[path];

            if (currentNode != null)
                property.SetValue(result, getObject(currentNode));
        }

        return result;
    }
    object? getObject(JsonNode node) => node.GetValue<JsonElement>().ValueKind switch
    {
        JsonValueKind.Null => null,
        JsonValueKind.Undefined => null,
        JsonValueKind.String => node.GetValue<string>(),
        JsonValueKind.Array => node.GetValue<JsonArray>(),
        JsonValueKind.Object => node.GetValue<JsonObject>(),
        JsonValueKind.Number => node.AsValue().TryGetValue(out int val) ? (object)val : (node.AsValue().TryGetValue(out float fval) ? (object)fval : (object)node.GetValue<double>()),
        JsonValueKind.False => false,
        JsonValueKind.True => true,
        _ => null
    };

    public override void Write(Utf8JsonWriter writer, T value, JsonSerializerOptions options) =>
        throw new NotImplementedException();
}

该方法可以让你在JsonPropertyName中指定json路径，也可以在不指定路径时使用字段名，并且支持自定义PropertyNamingPolicy（来自下面的解决方案）

用法：

class Model
{
    [JsonPropertyName("values.val")]
    public int ValuesVal { get; set; }

    [JsonPropertyName("single")]
    public int Value { get; set; }

    public string? OtherValue { get; set; }
}

string json = @"{""values"":{""val"":4}, ""single"": 2, ""other_value"": ""Here is other value""}";

// SnakeCaseNamingPolicy взят из решения  
// проблемы с политикой наименования ниже
// Данный параметр не обязателен, и нужен только
// для преобразования из "OtherValue" в "other_value"
Model? model = PathJsonConverter<Model>.Deserialize(json, new SnakeCaseNamingPolicy());

Console.WriteLine($"{model?.Value}, {model?.ValuesVal}, {model?.OtherValue}");

Newtonsoft.Json 的解决方案：

我建议您使用 Newtonsoft.Json，因为它提供了非常强大的 json 功能并且更可定制。

class JsonPathConverter : JsonConverter
{
    public override object ReadJson(JsonReader reader, Type objectType, object? existingValue, JsonSerializer serializer)
    {
        JObject jsonObj = JObject.Load(reader);
        object targetObj = Activator.CreateInstance(objectType) ?? throw new JsonException($"Can't create instance of {objectType}");

        foreach (PropertyInfo prop in objectType.GetProperties().Where(p => p.CanRead && p.CanWrite))
        {
            JsonPropertyAttribute? att = prop.GetCustomAttributes(true).OfType<JsonPropertyAttribute>().FirstOrDefault();
            string jsonPath = att?.PropertyName ?? prop.Name;
            JToken? token = jsonObj.SelectToken(jsonPath);

            if ((token?.Type ?? JTokenType.Null) != JTokenType.Null)
            {
                object? value = token?.ToObject(prop.PropertyType, serializer);
                prop.SetValue(targetObj, value, null);
            }
        }
        return targetObj;
    }

    public override bool CanConvert(Type objectType) => false;
    public override bool CanWrite => false;
    public override void WriteJson(JsonWriter writer, object? value, JsonSerializer serializer) => throw new NotImplementedException();
}

用法：

[JsonConverter(typeof(JsonPathConverter))]
class TestClass
{
    [JsonProperty("values.some_value")]
    public int SomeValue { get; set; }
}

TestClass testClass = JsonConvert.DeserializeObject<TestClass>(@"{""values"": {""some_value"": 5}}");

Console.WriteLine(testClass.SomeValue)

如果问题出在命名策略中：

如果问题是按照命名方法（如ObjectName和object_name），那么你就在这里

System.Net.Json的解决方案：

public class SnakeCaseNamingPolicy : JsonNamingPolicy
{
    public override string ConvertName(string name) => 
        string.Concat(name.Select((x, i) => i > 0 && char.IsUpper(x) ? "_" + x.ToString() : x.ToString())).ToLower();
}

用法：

TestObject testObject = JsonSerializer.Deserialize<TestObject>(
    "{\"int_value\": 5}", 
    new JsonSerializerOptions()
    {
        PropertyNamingPolicy = new SnakeCaseNamingPolicy()
    }
);
Console.WriteLine(testObject.IntValue);

Newtonsoft.Json 的解决方案：

此任务的解决方案已内置到库中：

TestObject testObject = JsonConvert.DeserializeObject<TestObject>(
    "{\"int_value\": 5}",
    new JsonSerializerSettings()
    {
        ContractResolver = new DefaultContractResolver()
        { NamingStrategy = new SnakeCaseNamingStrategy() }
    }
);
Console.WriteLine(testObject.IntValue);

找到了一个可能的解决方案 - 使用dynamic. 是的，最好不要滥用它，因为捕获状态更困难，但特别是在我的情况下，解决方案还不错。

        dynamic weatherData = JsonConvert.DeserializeObject(jsonData);
        weatherData = weatherData["current"];

        LastUpdate = weatherData["last_updated"];
        TemperatureCelsius = weatherData["temp_c"];
        FeelsLikeCelsius = weatherData["feelslike_c"];
        Discription = weatherData["condition"]["text"];
        WindSpeed = weatherData["wind_kph"];
        WindDirection = weatherData["wind_dir"];
        Pressure = weatherData["pressure_mb"];
        Precipitation = weatherData["precip_mm"];
        Humidity = weatherData["humidity"];
        Cloud = weatherData["cloud"];
        UvIndex = weatherData["uv"];
        IsDay = weatherData["is_day"];

更改：结果，我放弃了这样的解决方案，并按照评论中的建议创建了几个帮助器类。