YamlDotNet 根据值更改属性名称答案

【问题标题】：YamlDotNet change property name according to valueYamlDotNet 根据值更改属性名称
【发布时间】：2021-01-18 18:00:49
【问题描述】：

我正在尝试根据我在 yaml 文件的输出中获得的值来格式化对象的属性名称。我正在使用 yamldotnet 库。以下是我的课程

State.cs

public class State
    {
        public string name { get; set; }
        public List<Operation> operations { get; set; }

    }

Operation.cs

public class Operation
    {
        public string name { get; set; }
        public string type { get; set; }
    }

在状态类中，我有操作列表。序列化后，我得到以下 yaml 输出：

states:
- name: uPError
  operations:
  - name: switch on uP
    type: entry
  - name: switch off uP
    type: exit
  - name: test Do
    type: do

但我打算显示的格式如下：

states:
- name: uPError
  entryActions: [switch on uP]
  exitActions: [switch off uP]
  doActions: [test Do]

我正在获取类型及其名称，根据类型，我需要将其更改为 entryAction、exitActions 或 doActions 和然后附加相应的名称。如何实现上述预期格式？提前致谢。

【问题讨论】：

标签： c# .net yaml yamldotnet

【解决方案1】：

实现这一点的一种方法是实现IYamlTypeConverter 来自定义State 类型的转换。这使您可以控制类的整个序列化：

public class StateConverter : IYamlTypeConverter
{
    public bool Accepts(Type type)
    {
        return type == typeof(State);
    }

    public object? ReadYaml(IParser parser, Type type)
    {
        var state = new State { operations = new List<UserQuery.Operation>() };
        
        parser.Consume<MappingStart>();
        while (!parser.TryConsume<MappingEnd>(out var _))
        {
            var key = parser.Consume<Scalar>().Value;
            if (key == nameof(state.name))
            {
                state.name = parser.Consume<Scalar>().Value;
            }
            else
            {
                parser.Consume<SequenceStart>();
                while (!parser.TryConsume<SequenceEnd>(out var _))
                {
                    state.operations.Add(new Operation
                    {
                       type = key,
                       name = parser.Consume<Scalar>().Value,
                    });
                }
            }
        }
        
        return state;
    }

    public void WriteYaml(IEmitter emitter, object? value, Type type)
    {
        var state = (State)value!;
        
        emitter.Emit(new MappingStart());
        
        emitter.Emit(new Scalar(nameof(state.name)));
        emitter.Emit(new Scalar(state.name));

        foreach (var operationGroup in state.operations.GroupBy(o => o.type))
        {
            emitter.Emit(new Scalar(operationGroup.Key));

            emitter.Emit(new SequenceStart(null, null, false, SequenceStyle.Flow));
            foreach (var operation in operationGroup)
            {
                emitter.Emit(new Scalar(operation.name));
            }
            emitter.Emit(new SequenceEnd());
        }

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

您需要在SerializerBuilder 和DeserializerBuilder 上注册该类型：

var serializer = new SerializerBuilder()
    .WithTypeConverter(new StateConverter())
    .Build();

var deserializer = new DeserializerBuilder()
    .WithTypeConverter(new StateConverter())
    .Build();

【讨论】：