Question

我需要序列化包含具有getter但没有setter的属性的对象（OpenTK.Vector2）。我希望一般可以忽略这些属性，否则我最终会从具有两个相关数据片段（X和Y）的对象中充分夸大JSON。

代码：

JsonSerializerSettings settings = new JsonSerializerSettings { ReferenceLoopHandling = ReferenceLoopHandling.Ignore };
Vector2 v = new Vector2 { X = 1, Y = 0 };
string json = JsonConvert.SerializeObject(v, settings);

生成字符串：

{
   "X" : 1.0,
   "Y" : 0.0,
   "Length" : 1.0,
   "LengthFast" : 1.0016948,
   "LengthSquared" : 1.0,
   "PerpendicularRight" : {
      "X" : 0.0,
      "Y" : -1.0,
      "Length" : 1.0,
      "LengthFast" : 1.0016948,
      "LengthSquared" : 1.0,
      "PerpendicularRight" : {
         "X" : -1.0,
         "Y" : 0.0,
         "Length" : 1.0,
         "LengthFast" : 1.0016948,
         "LengthSquared" : 1.0,
         "PerpendicularRight" : {
            "X" : 0.0,
            "Y" : 1.0,
            "Length" : 1.0,
            "LengthFast" : 1.0016948,
            "LengthSquared" : 1.0
         }
      },
      "Yx" : {
         "X" : -1.0,
         "Y" : 0.0,
         "Length" : 1.0,
         "LengthFast" : 1.0016948,
         "LengthSquared" : 1.0,
         "PerpendicularRight" : {
            "X" : 0.0,
            "Y" : 1.0,
            "Length" : 1.0,
            "LengthFast" : 1.0016948,
            "LengthSquared" : 1.0
         } 
      }
   },
   "PerpendicularLeft" : {
      "X" : 0.0,
      "Y" : 1.0,
      "Length" : 1.0,
      "LengthFast" : 1.0016948,
      "LengthSquared" : 1.0,
      "PerpendicularLeft" : {
         "X" : -1.0,
         "Y" : 0.0,
         "Length" : 1.0,
         "LengthFast" : 1.0016948,
         "LengthSquared" : 1.0,
         "PerpendicularLeft" : {
            "X" : 0.0,
            "Y" : -1.0,
            "Length" : 1.0,
            "LengthFast" : 1.0016948,
            "LengthSquared" : 1.0
         },
         "Yx" : {
            "X" : 0.0,
            "Y" : -1.0,
            "Length" : 1.0,
            "LengthFast" : 1.0016948,
            "LengthSquared" : 1.0
         }
      }
   },
   "Yx" : {
      "X" : 0.0,
      "Y" : 1.0,
      "Length" : 1.0,
      "LengthFast" : 1.0016948,
      "LengthSquared" : 1.0,
      "PerpendicularLeft" : {
         "X" : -1.0,
         "Y" : 0.0,
         "Length" : 1.0,
         "LengthFast" : 1.0016948,
         "LengthSquared" : 1.0,
         "PerpendicularLeft" : {
            "X" : 0.0,
            "Y" : -1.0,
            "Length" : 1.0,
            "LengthFast" : 1.0016948,
            "LengthSquared" : 1.0
         },
         "Yx" : {
            "X" : 0.0,
            "Y" : -1.0,
            "Length" : 1.0,
            "LengthFast" : 1.0016948,
            "LengthSquared" : 1.0
         }
      }
   }
}

如何让序列化程序忽略这些其他属性？

Answer 1

由于您无法修改OpenTK.Vector2结构以将[JsonIgnore]属性添加到get-only属性，因此最简单的方法可能是编写自己的JsonConverter为它：

public class Vector2Converter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return objectType == typeof(OpenTK.Vector2);
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        var temp = JObject.Load(reader);
        return new OpenTK.Vector2(((float?)temp["X"]).GetValueOrDefault(), ((float?)temp["Y"]).GetValueOrDefault());
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        var vec = (OpenTK.Vector2)value;
        serializer.Serialize(writer, new { X = vec.X, Y = vec.Y});
    }
}

然后使用它：

        var settings = new JsonSerializerSettings();
        settings.Converters.Add(new Vector2Converter());
        Vector2 v = new Vector2 { X = 1, Y = 0 };
        string json = JsonConvert.SerializeObject(v, settings);
        Debug.WriteLine(json);

哪个产生

{"X":1.0,"Y":0.0}

但是如果你真的想忽略所有类和结构上的所有get-only属性（可能会产生无法预料的后果），请参见：Is there a way to ignore get-only properties in Json.NET without using JsonIgnore attributes?。

不要在没有setter的情况下序列化属性

1 个答案: