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Type is an interface or abstract

Type is an interface or abstract

作者: 天山的虫 | 来源:发表于2019-10-25 15:23 被阅读0次

抽象类或者接口:

public abstract class BaseFoo
{
    public string FooBarBuzz { get; set; }
}

public class AFoo : BaseFoo
{
    public string A { get; set; }
}

public class BFoo : BaseFoo
{
    public string B { get; set; }
}

构造JSON内容

AFoo a = new AFoo();
a.FooBarBuzz = "A";
a.A = "Hello World";

BFoo b = new BFoo();
b.FooBarBuzz = "B";
b.B = "Hello World";

List<BaseFoo> allFoos = new List<BaseFoo>();
allFoos.Add(a);
allFoos.Add(b);

var result = JsonConvert.SerializeObject(allFoos);

结果值

Json值

异常代码:

var test = JsonConvert.DeserializeObject<List<BaseFoo>>(result);

问题

抽象类序列化失败

解决方案:

1.构造相应的转化类

注意ReadJson方法,根据不同的类型,进行不同的序列化

public class FooConverter : JsonConverter
{
    public override bool CanConvert(Type objectType)
    {
        return (objectType == typeof(BaseFoo));
    }

    public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
    {
        JObject jo = JObject.Load(reader);
        if (jo["FooBarBuzz"].Value<string>() == "A")
            return jo.ToObject<AFoo>(serializer);

        if (jo["FooBarBuzz"].Value<string>() == "B")
            return jo.ToObject<BFoo>(serializer);

        return null;
    }

    public override bool CanWrite
    {
        get { return false; }
    }

    public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
    {
        throw new NotImplementedException();
    }
}

2.使用

JsonConverter[] converters = { new FooConverter()};
var test = JsonConvert.DeserializeObject<List<BaseFoo>>(result, new JsonSerializerSettings() { Converters = converters });

引用:

1.英文原文

2.stackoverflow帖链接

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