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YamlDotNet -自定义序列化
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Stack Overflow用户
提问于 2020-10-07 18:20:24
回答 1查看 1.1K关注 0票数 1

我有一个表示RPC方法调用的.NET类,如下所示:

代码语言:javascript
复制
class MethodCall
{
    public string MethodName { get; set; }
    public Collection<object> Arguments { get; set; }
}

我想将一个Collection<MethodCall>序列化为YAML。我使用YamlDotNet来实现这一点。

默认情况下,YamlDotNet将序列化这些对象,如下所示:

代码语言:javascript
复制
methodName: someName
arguments:
- arg1
- arg2
- ...

我想将生成的YAML简化为:

代码语言:javascript
复制
someName:
- arg1
- arg2

有什么简单的方法可以做到这一点吗?请注意,参数可以是复杂的对象(即不是简单的标量)。

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-10-08 18:17:24

您可以通过注册一个执行所需转换的IYamlTypeConverter实现来实现此功能。

下面是一个可能的实现:

代码语言:javascript
复制
public sealed class MethodCallConverter : IYamlTypeConverter
{
    // Unfortunately the API does not provide those in the ReadYaml and WriteYaml
    // methods, so we are forced to set them after creation.
    public IValueSerializer ValueSerializer { get; set; }
    public IValueDeserializer ValueDeserializer { get; set; }
    
    public bool Accepts(Type type) => type == typeof(MethodCall);

    public object ReadYaml(IParser parser, Type type)
    {
        parser.Consume<MappingStart>();

        var call = new MethodCall
        {
            MethodName = (string)ValueDeserializer.DeserializeValue(parser, typeof(string), new SerializerState(), ValueDeserializer),
            Arguments = (Collection<object>)ValueDeserializer.DeserializeValue(parser, typeof(Collection<object>), new SerializerState(), ValueDeserializer),
        };

        parser.Consume<MappingEnd>();
        
        return call;
    }
    
    public void WriteYaml(IEmitter emitter, object value, Type type)
    {
        emitter.Emit(new MappingStart());

        var call = (MethodCall)value;
        ValueSerializer.SerializeValue(emitter, call.MethodName, typeof(string));
        ValueSerializer.SerializeValue(emitter, call.Arguments, typeof(Collection<object>));

        emitter.Emit(new MappingEnd());
    }
}

转换器需要通过WithTypeConverter方法注册到SerializerBuilderDeserializerBuilder中。请注意,YamlDotNet没有为我们提供递归调用(反)序列化程序的方法,因此我们必须设置一些公共属性作为解决办法。这并不像它可能的那样干净,但仍然有效:

代码语言:javascript
复制
string SerializeMethodCall(MethodCall call)
{
    var methodCallConverter = new MethodCallConverter();
    var serializerBuilder = new SerializerBuilder()
        .WithNamingConvention(CamelCaseNamingConvention.Instance)
        .WithTypeConverter(methodCallConverter);

    methodCallConverter.ValueSerializer = serializerBuilder.BuildValueSerializer();

    var serializer = serializerBuilder.Build();

    var yaml = serializer.Serialize(call);
    return yaml;
}

MethodCall DeserializeMethodCall(string yaml)
{
    var methodCallConverter = new MethodCallConverter();
    var deserializerBuilder = new DeserializerBuilder()
        .WithNamingConvention(CamelCaseNamingConvention.Instance)
        .WithTypeConverter(methodCallConverter);

    methodCallConverter.ValueDeserializer = deserializerBuilder.BuildValueDeserializer();

    var deserializer = deserializerBuilder.Build();
    var call = deserializer.Deserialize<MethodCall>(yaml);
    return call;
}
票数 2
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/64242023

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