在.NET中序列化大量链接的数据(自定义JSON.NET引用)

Question

我希望在序列化数据时避免重新发明轮子.我知道一些方法来序列化彼此链接的对象,但它的范围从编写一些代码到编写大量用于序列化的代码,我想避免这种情况.必须有一些通用的解决方案.

假设我有这样的结构:

Person
    bro = new Person { name = "bro", pos = new Pos { x = 1, y = 5 } },
    sis = new Person { name = "sis", pos = new Pos { x = 2, y = 6 } },
    mom = new Person { name = "mom", pos = new Pos { x = 3, y = 7 }, 
        children = new List<Person> { bro, sis }
    },
    dad = new Person { name = "dad", pos = new Pos { x = 4, y = 8 }, 
        children = new List<Person> { bro, sis }, mate = mom
    };
mom.mate = dad;
Family family = new Family { persons = new List<Person> { mom, dad, bro, sis } };

我想将数据序列化为这样的东西:

family: {
    persons: [
        { name: "bro", pos: { x: 1, y: 5 } },
        { name: "sis", pos: { x: 2, y: 6 } },
        { name: "mom", pos: { x: 3, y: 7 }, mate: "dad", children: [ "bro", "sis" ] },
        { name: "dad", pos: { x: 4, y: 8 }, mate: "mom", children: [ "bro", "sis" ] },
    ]
}

在这里,链接被序列化为名称,假设名称是唯一的.链接也可以是"family.persons.0"或生成的唯一ID或其他.

要求:

1. 格式必须是人类可读的,并且最好是人类可写的.因此,按优先顺序排列:JSON,YAML*,XML,自定义.没有二进制格式.

2. 序列化必须支持.NET提供的所有好东西.泛型是必须的,包括IEnumerable <>,IDictionary <>等类型.动态类型/无类型对象是可取的.

3. 格式不能是可执行的.没有Lua,Python等脚本和类似的东西.

4. 如果生成唯一ID,则它们必须是稳定的(通过序列化 - 反序列化保持),因为文件将被放入版本控制系统中.

*听说过YAML,可悲的是,它似乎已经死了.

Answer 1

使用JSON.NET解决了这个问题(很棒的库!).现在,对象首先被序列化并准确引用我想要的对象; 第二,没有多少"$ id"和"$ ref"字段.在我的解决方案中,对象的第一个属性用作其标识符.

我创建了两个JsonConvertors(用于引用对象和引用的对象):

interface IJsonLinkable
{
    string Id { get; }
}

class JsonRefConverter : JsonConverter
{
    public override void WriteJson (JsonWriter writer, object value, JsonSerializer serializer)
    {
        writer.WriteValue(((IJsonLinkable)value).Id);
    }

    public override object ReadJson (JsonReader reader, Type type, object existingValue, JsonSerializer serializer)
    {
        if (reader.TokenType != JsonToken.String)
            throw new Exception("Ref value must be a string.");
        return JsonLinkedContext.GetLinkedValue(serializer, type, reader.Value.ToString());
    }

    public override bool CanConvert (Type type)
    {
        return type.IsAssignableFrom(typeof(IJsonLinkable));
    }
}

class JsonRefedConverter : JsonConverter
{
    public override void WriteJson (JsonWriter writer, object value, JsonSerializer serializer)
    {
        serializer.Serialize(writer, value);
    }

    public override object ReadJson (JsonReader reader, Type type, object existingValue, JsonSerializer serializer)
    {
        var jo = JObject.Load(reader);
        var value = JsonLinkedContext.GetLinkedValue(serializer, type, (string)jo.PropertyValues().First());
        serializer.Populate(jo.CreateReader(), value);
        return value;
    }

    public override bool CanConvert (Type type)
    {
        return type.IsAssignableFrom(typeof(IJsonLinkable));
    }
}

以及保存引用数据的上下文(每个类型都有一个字典,因此ID只能在相同类型的对象中唯一):

class JsonLinkedContext
{
    private readonly IDictionary<Type, IDictionary<string, object>> links = new Dictionary<Type, IDictionary<string, object>>();

    public static object GetLinkedValue (JsonSerializer serializer, Type type, string reference)
    {
        var context = (JsonLinkedContext)serializer.Context.Context;
        IDictionary<string, object> links;
        if (!context.links.TryGetValue(type, out links))
            context.links[type] = links = new Dictionary<string, object>();
        object value;
        if (!links.TryGetValue(reference, out value))
            links[reference] = value = FormatterServices.GetUninitializedObject(type);
        return value;
    }
}

属性的一些属性是必要的:

[JsonObject(MemberSerialization.OptIn)]
class Family
{
    [JsonProperty(ItemConverterType = typeof(JsonRefedConverter))]
    public List<Person> persons;
}

[JsonObject(MemberSerialization.OptIn)]
class Person : IJsonLinkable
{
    [JsonProperty]
    public string name;
    [JsonProperty]
    public Pos pos;
    [JsonProperty, JsonConverter(typeof(JsonRefConverter))]
    public Person mate;
    [JsonProperty(ItemConverterType = typeof(JsonRefConverter))]
    public List<Person> children;

    string IJsonLinkable.Id { get { return name; } }
}

[JsonObject(MemberSerialization.OptIn)]
class Pos
{
    [JsonProperty]
    public int x;
    [JsonProperty]
    public int y;
}

所以,当我使用这段代码序列化和反序列化时:

JsonConvert.SerializeObject(family, Formatting.Indented, new JsonSerializerSettings {
    NullValueHandling = NullValueHandling.Ignore,
    Context = new StreamingContext(StreamingContextStates.All, new JsonLinkedContext()),
});

JsonConvert.DeserializeObject<Family>(File.ReadAllText(@"..\..\Data\Family.json"), new JsonSerializerSettings {
    Context = new StreamingContext(StreamingContextStates.All, new JsonLinkedContext()),
});

我得到了这个整洁的JSON:

{
  "persons": [
    {
      "name": "mom",
      "pos": {
        "x": 3,
        "y": 7
      },
      "mate": "dad",
      "children": [
        "bro",
        "sis"
      ]
    },
    {
      "name": "dad",
      "pos": {
        "x": 4,
        "y": 8
      },
      "mate": "mom",
      "children": [
        "bro",
        "sis"
      ]
    },
    {
      "name": "bro",
      "pos": {
        "x": 1,
        "y": 5
      }
    },
    {
      "name": "sis",
      "pos": {
        "x": 2,
        "y": 6
      }
    }
  ]
}

在我的解决方案中,我不喜欢的是我必须使用JObject,即使从技术上讲它是不必要的.它可能会创建相当多的对象,因此加载速度会变慢.但看起来这是最常用的自定义对象转换器的方法.无论如何,可用于避免这种情况的方法都是私有的.