Json.net序列化特定的私有字段

时间:2015-08-14 11:22:22

标签: c# serialization json.net

我有以下课程:

public class TriGrid
{
    private List<HexTile> _hexes;
    //other private fields...
    //other public proprerties
}

我的目标是仅序列化_hexes字段,因此我创建了以下ContractResolver:

internal class TriGridContractResolver : DefaultContractResolver
{
    protected override List<MemberInfo> GetSerializableMembers(Type objectType)
    {
        return new List<MemberInfo> { objectType.GetMember("_hexes", BindingFlags.NonPublic | BindingFlags.Instance)[0] };
    }
}

当我想序列化TriGrid的一个实例时,我做了:

        var settings = new JsonSerializerSettings()
        {
            ContractResolver = new TriGridContractResolver()
        };
        var json = JsonConvert.SerializeObject(someTriGrid, settings);
        string strintJson = json.ToString();

但是当我检查strintJson的值始终为"{}"时。 _hexes有元素,不是空的。如果我序列化一个特定的HexTile它按预期工作。我在这里做错了什么?

2 个答案:

答案 0 :(得分:63)

无需实施自定义DefaultContractResolver。解决方案是将[JsonProperty]放在_hexes上,[JsonIgnore]放在所有其他属性和字段上。

答案 1 :(得分:5)

因为商业模式最终会发生变化,所以我更喜欢实施ISerializable并使用.NET创建游戏的方式(即属性包)。当您需要在运行时对对象进行版本化时,这最有效。任何你不想序列化的东西,都不要把它放在财产袋里。

特别是,因为JSON.Net(Newtonsoft.Json)也会通过序列化和反序列化方法来表达它。

using System;
using System.Runtime.Serialization;

[Serializable]
public class Visitor : ISerializable
{
    private int Version;

    public string Name { get; private set; }

    public string IP { get; set: }

    public Visitor()
    {
        this.Version = 2;
    }

    public void ChangeName(string Name)
    {
        this.Name = Name;
    }

    //Deserialize
    protected Visitor(SerializationInfo info, StreamingContext context)
    {
        this.Version = info.GetInt32("Version");
        this.Name = info.GetString("Name");
    }

    //Serialize
    public void GetObjectData(SerializationInfo info, StreamingContext context)
    {
        info.AddValue("Version", this.Version);

        info.AddValue("Name", this.Name);
    }

    [OnDeserialized]
    private void OnDeserialization(StreamingContext context)
    {
        switch (this.Version)
        {
            case 1:
                //Handle versioning issues, if this
                //deserialized version is one, so that
                //it can play well once it's serialized as
                //version two.
                break;
        }
    }
}