Json.net 仅序列化某些属性 [英] Json.net serialize only certain properties

问题描述

Json.net 有没有办法只指定你想要序列化的属性?或者基于像 Declared Only 这样的绑定标志序列化某些属性?

Does Json.net have any way to specify only the properties you want to be serialized? or alternatively serialize certain properties based on binding flags like Declared Only?

现在我正在使用 JObject.FromObject(MainObj.SubObj); 来获取 SubObj 的所有属性，它是一个遵循 ISubObject 接口的类的实例:

Right now I am using JObject.FromObject(MainObj.SubObj); to get all properties of SubObj which is an instance of a class that obeys the ISubObject interface:

public interface ISubObject
{

}

public class ParentSubObject : ISubObject
{
    public string A { get; set; }
}


public class SubObjectWithOnlyDeclared : ParentSubObject
{
    [JsonInclude] // This is fake, but what I am wishing existed
    public string B { get; set; }

    [JsonInclude] // This is fake, but what I am wishing existed
    public string C { get; set; }
}

public class NormalSubObject: ParentSubObject
{
    public string B { get; set; }
}

如果 MainObj.SubObj 是 NormalSubObject 它将同时对 A 和 B 进行序列化，但如果它是 SubObjectWithOnlyDeclared 它只会对 B 和 C 进行序列化并忽略父属性

If MainObj.SubObj was a NormalSubObject it would serailize both A and B but if it was SubObjectWithOnlyDeclared it would serailize only B and C and ignore the parent property

推荐答案

你可以像下面这样写一个自定义的ContractResolver

You can write a custom ContractResolver like below

public class IgnoreParentPropertiesResolver : DefaultContractResolver
{
    bool IgnoreBase = false;
    public IgnoreParentPropertiesResolver(bool ignoreBase)
    {
        IgnoreBase = ignoreBase;
    }
    protected override IList<JsonProperty> CreateProperties(Type type, MemberSerialization memberSerialization)
    {
        var allProps = base.CreateProperties(type, memberSerialization);
        if (!IgnoreBase) return allProps;

        //Choose the properties you want to serialize/deserialize
        var props = type.GetProperties(~BindingFlags.FlattenHierarchy); 

        return allProps.Where(p => props.Any(a => a.Name == p.PropertyName)).ToList();
    }
}

<小时>

现在您可以在序列化过程中将其用作:

---

Now you can use it in your serialization process as:

var settings = new JsonSerializerSettings() { 
                      ContractResolver = new IgnoreParentPropertiesResolver(true) 
               };
var json1 = JsonConvert.SerializeObject(new SubObjectWithOnlyDeclared(),settings );

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