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c# - 基于 JsonSerializerSettings 确定 JSON.Net 属性映射

转载 作者:行者123 更新时间:2023-11-30 17:30:54 24 4
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给定一个 JsonSerializerSettings 对象、一个类型和一个 System.Reflection.PropertyInfo 对象,我如何确定:

1) 当使用给定设置使用 JSON.Net 转换时,该属性的 JSON 属性名称是什么?鉴于属性或类本身可能存在或不存在的各种规则和属性,以及可能在 JsonSerializerSettings 中设置的各种设置?

2) 在使用 JSON.Net 进行反序列化时,如何知道给定的“JSON”属性名称是否会映射到 c# 属性?如果是,是哪个属性?

class Person
{
[JsonProperty("person_id")]
Guid PersonId { get; set; } // given property info for this, want to recieve "person_id"

string FirstName { get; set; } // will this serialize to "firstName"? "FirstName"? depending on the settings?
string LastName { get; set; } //given the settings, would { "LASTNAME": "Johnson" } be serialized into this property?

[JsonIgnore]
string SSN { get; set; } // determine will this property be mapped?
}

请注意,我希望找到一种解决方案,而不是自己查看属性属性(考虑到所有可能性和契约(Contract)解析器,这太容易出错),我想根据 JsonSerializerSettings 确定属性映射是什么?

最佳答案

您需要的信息可从 Json.NET 的 contract resolver 获得。 .访问它的步骤顺序如下:

  1. 制造 JsonSerializer从设置。
  2. 获取其 contract resolver .
  3. 解决您的契约(Contract) Person输入并将其转换为 JsonObjectContract .
  4. 找到 JsonProperty Properties 为您的属性(property)列表。有关该特定属性的所有序列化信息都在那里。

因此,例如您可以创建以下扩展方法:

public static partial class JsonExtensions
{
static JsonProperty GetProperty(this JsonSerializerSettings settings, Type type, string underlyingName)
{
// Use JsonSerializer.Create(settings) instead if your framework ignores the global JsonConvert.DefaultSettings
var resolver = JsonSerializer.CreateDefault(settings).ContractResolver;
var contract = resolver.ResolveContract(type) as JsonObjectContract;
if (contract == null)
throw new ArgumentException(string.Format("{0} is not a JSON object", type));
return contract.Properties.Where(p => p.UnderlyingName == underlyingName).SingleOrDefault();
}

public static string GetPropertyName(this JsonSerializerSettings settings, Type type, string underlyingName)
{
var property = settings.GetProperty(type, underlyingName);
// The property might be null if it is nonpublic and not marked with [JsonProperty]
return property == null ? null : property.PropertyName;
}

public static bool GetIsIgnored(this JsonSerializerSettings settings, Type type, string underlyingName)
{
var property = settings.GetProperty(type, underlyingName);
// The property might be null if it is nonpublic and not marked with [JsonProperty]
return property == null ? true : property.Ignored;
}
}

那么,如果你使用如下方法:

var settings = new JsonSerializerSettings
{
ContractResolver = new CamelCasePropertyNamesContractResolver(),
};

foreach (var property in typeof(Person).GetProperties(BindingFlags.Public | BindingFlags.NonPublic | BindingFlags.Instance))
{
Console.WriteLine("Property {0}: Json name = \"{1}\", IsIgnored = {2}", property, settings.GetPropertyName(typeof(Person), property.Name), settings.GetIsIgnored(typeof(Person), property.Name));
}

输出将是

Property System.Guid PersonId: Json name = "person_id", IsIgnored = False
Property System.String FirstName: Json name = "firstName", IsIgnored = False
Property System.String LastName: Json name = "lastName", IsIgnored = False
Property System.String SSN: Json name = "ssn", IsIgnored = True

注意事项:

  • Json.NET 维护全局默认 JsonSerializerSettings可以通过 JsonConvert.DefaultSettings 访问.这个属性

    Gets or sets a function that creates default JsonSerializerSettings. Default settings are automatically used by serialization methods on JsonConvert, and ToObject<T> () and FromObject(Object) on JToken. To serialize without using any default settings create a JsonSerializer with Create().

    如果您的框架忽略了全局默认设置,您应该替换 JsonSerializer.CreateDefault(settings)JsonSerializer.Create(settings)在上述方法中。

  • 为了提高性能,您可以一次制造序列化程序并将扩展方法添加到 JsonSerializer而不是 JsonSerializerSettings .如果你需要一个给定属性的多个属性,你可以只获取 JsonProperty并直接访问它。

  • Json.NET 可以序列化字段和属性,因此需要 PropertyInfo传递给扩展方法可能会限制对某些必要序列化信息的访问。

sample 加工 .Net fiddle .

关于c# - 基于 JsonSerializerSettings 确定 JSON.Net 属性映射,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48937615/

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