Question

我有一个（简化的）对象，像这样：

public class Contact 
{
    [Ignore]
    public string Name { get; set;}
}

然后我对对象进行序列化和反序列化，以创建深度复制克隆（https://stackoverflow.com/a/78612/2987066），如下所示：

var deserializeSettings = new JsonSerializerSettings 
{
    ObjectCreationHandling = ObjectCreationHandling.Replace,
};

var clone = JsonConvert.SerializeObject(contact, deserializeSettings);

return JsonConvert.DeserializeObject<T>(clone);

但该属性不会被复制

是否可以序列化标记为[Ignore]的属性。也许使用JsonSerializerSettings还是我需要将其标记为[JSONProperty]

Answer 1

我想做到这一点的一种方法是重新添加被忽略的字段：

JsonConvert.DeserializeObject<Contact>(clone).Name=contact.Name

但是使用ignore的目的是使它不会被序列化。

另一个选择是在jsonserializersettings中指定自定义合同解析器：

 var deserializeSettings = new JsonSerializerSettings
            {
                ObjectCreationHandling = ObjectCreationHandling.Replace,
                ContractResolver = new DynamicContractResolver()
            };

https://www.newtonsoft.com/json/help/html/T_Newtonsoft_Json_Serialization_IContractResolver.htm

public class DynamicContractResolver: DefaultContractResolver
{
    protected override JsonProperty CreateProperty(MemberInfo member, MemberSerialization memberSerialization)
    {
        JsonProperty property = base.CreateProperty(member, memberSerialization);

        //property.HasMemberAttribute = true;
        property.Ignored = false;

        //property.ShouldSerialize = instance =>
        //{
        //    return true;
        //};

        return property;
    }
}

是否可以对JSON自动序列化标记为[忽略]的属性

1 个答案: