c# - 如何在自定义 System.Text.Json JsonConverter 中使用默认序列化？

我正在写一个 custom System.Text.Json.JsonConverter将旧数据模型升级到新版本。我已覆盖 Read()并实现了必要的后处理。但是，我根本不需要在 Write()中做任何自定义操作。方法。如果我根本没有转换器，如何自动生成默认序列化？显然我可以使用不同的 JsonSerializerOptions用于反序列化和序列化，但是我的框架并没有直接为每个提供不同的选项。
下面是一个简化的例子。假设我以前有以下数据模型:

public record Person(string Name);

我已经升级到

public record Person(string FirstName, string LastName);

我写了一个转换器如下:

public sealed class PersonConverter : JsonConverter<Person>
{
    record PersonDTO(string FirstName, string LastName, string Name); // A DTO with both the old and new properties.

    public override Person Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options)
    {
        var dto = JsonSerializer.Deserialize<PersonDTO>(ref reader, options);
        var oldNames = dto?.Name?.Split(' ', StringSplitOptions.RemoveEmptyEntries) ?? Enumerable.Empty<string>();
        return new Person(dto.FirstName ?? oldNames.FirstOrDefault(), dto.LastName ?? oldNames.LastOrDefault());
    }

    public override void Write(Utf8JsonWriter writer, Person person, JsonSerializerOptions options)
        => // What do I do here? I want to preserve other options such as options.PropertyNamingPolicy, which are lost by the following call
        JsonSerializer.Serialize(writer, person);
}

和往返

var options = new JsonSerializerOptions
{
    PropertyNamingPolicy = JsonNamingPolicy.CamelCase,
    Converters = { new PersonConverter() },
};
var person = JsonSerializer.Deserialize<Person>(json, options);
var json2 = JsonSerializer.Serialize(person, options);

那么结果是{"FirstName":"FirstName","LastName":"LastName"} -- 即序列化期间的驼峰式 shell 丢失。但是如果我在编写时通过递归调用传递选项

    public override void Write(Utf8JsonWriter writer, Person person, JsonSerializerOptions options)
        => // What do I do here? I want to preserve other options such as options.PropertyNamingPolicy, which are lost by the following call
        JsonSerializer.Serialize(writer, person, options);

然后序列化因堆栈溢出而失败。
如何获得忽略自定义转换器的精确默认序列化？没有等效于 Json.NET 的 JsonConverter.CanWrite属性(property)。
演示 fiddle here .

最佳答案

如 docs 中所述, 转换器的选择具有以下优先级:

• [JsonConverter] applied to a property.
• A converter added to the Converters collection.
• [JsonConverter] applied to a custom value type or POCO.

每个案例都需要单独处理。

• 如果您有 [JsonConverter]应用于属性。然后只需调用 JsonSerializer.Serialize(writer, person, options);将生成默认序列化。
• 如果您将 A 转换器添加到 Converters集合。然后在 Write() 里面(或 Read()) 方法，可以复制传入的 options使用 JsonSerializerOptions copy constructor , 从副本的 Converters中删除转换器列表，并将修改后的副本传递给 JsonSerializer.Serialize<T>(Utf8JsonWriter, T, JsonSerializerOptions);这在 .NET Core 3.x 中无法轻松完成，因为该版本中不存在复制构造函数。临时修改Converters移除转换器的传入选项的集合不是线程安全的，因此不推荐。相反，需要创建新选项并手动复制每个属性以及 Converters集合，跳过 converterType 类型的转换.
• 如果您有 [JsonConverter]应用于自定义值类型或 POCO。似乎没有生成默认序列化的方法。

因为，在问题中，转换器被添加到 Converters列表，以下修改版本正确生成默认序列化:

public sealed class PersonConverter : DefaultConverterFactory<Person>
{
    record PersonDTO(string FirstName, string LastName, string Name); // A DTO with both the old and new properties.

    protected override Person Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions modifiedOptions, JsonConverter<Person> defaultConverter)
    {
        var dto = JsonSerializer.Deserialize<PersonDTO>(ref reader, modifiedOptions);
        var oldNames = dto?.Name?.Split(' ', StringSplitOptions.RemoveEmptyEntries) ?? Enumerable.Empty<string>();
        return new Person(dto.FirstName ?? oldNames.FirstOrDefault(), dto.LastName ?? oldNames.LastOrDefault());
    }
}

public abstract class DefaultConverterFactory<T> : JsonConverterFactory
{
    class DefaultConverter : JsonConverter<T>
    {
        readonly JsonSerializerOptions modifiedOptions;
        readonly DefaultConverterFactory<T> factory;
        readonly JsonConverter<T> defaultConverter;

        public DefaultConverter(JsonSerializerOptions options, DefaultConverterFactory<T> factory)
        {
            this.factory = factory;
            this.modifiedOptions = options.CopyAndRemoveConverter(factory.GetType());
            this.defaultConverter = (JsonConverter<T>)modifiedOptions.GetConverter(typeof(T));
        }

        public override void Write(Utf8JsonWriter writer, T value, JsonSerializerOptions options) => factory.Write(writer, value, modifiedOptions, defaultConverter);

        public override T Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options) => factory.Read(ref reader, typeToConvert, modifiedOptions, defaultConverter);
    }

    protected virtual T Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions modifiedOptions, JsonConverter<T> defaultConverter)
        => defaultConverter.ReadOrSerialize<T>(ref reader, typeToConvert, modifiedOptions);

    protected virtual void Write(Utf8JsonWriter writer, T value, JsonSerializerOptions modifiedOptions, JsonConverter<T> defaultConverter)
        => defaultConverter.WriteOrSerialize(writer, value, modifiedOptions);

    public override bool CanConvert(Type typeToConvert) => typeof(T) == typeToConvert;

    public override JsonConverter CreateConverter(Type typeToConvert, JsonSerializerOptions options) => new DefaultConverter(options, this);
}

public static class JsonSerializerExtensions
{
    public static JsonSerializerOptions CopyAndRemoveConverter(this JsonSerializerOptions options, Type converterType)
    {
        var copy = new JsonSerializerOptions(options);
        for (var i = copy.Converters.Count - 1; i >= 0; i--)
            if (copy.Converters[i].GetType() == converterType)
                copy.Converters.RemoveAt(i);
        return copy;
    }

    public static void WriteOrSerialize<T>(this JsonConverter<T> converter, Utf8JsonWriter writer, T value, JsonSerializerOptions options)
    {
        if (converter != null)
            converter.Write(writer, value, options);
        else
            JsonSerializer.Serialize(writer, value, options);
    }

    public static T ReadOrSerialize<T>(this JsonConverter<T> converter, ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options)
    {
        if (converter != null)
            return converter.Read(ref reader, typeToConvert, options);
        else
            return (T)JsonSerializer.Deserialize(ref reader, typeToConvert, options);
    }
}

笔记:

• 我使用转换器工厂而不是转换器作为 PersonConverter 的基类因为它让我可以方便地在制造的转换器中缓存复制的选项和默认转换器。
• 如果您尝试申请 DefaultConverterFactory<T>到自定义值类型或 POCO，例如

  [JsonConverter(typeof(PersonConverter))] public record Person(string FirstName, string LastName);
  

  将发生令人讨厌的堆栈溢出。

演示 fiddle here .

关于c# - 如何在自定义 System.Text.Json JsonConverter 中使用默认序列化？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/65430420/