自定义JSON序列化序列化和无视类属性反序列化的所有属性(Custom Json Serialize

我想我的序列化类的所有属性，但要隐藏某些属性，同时返回响应。

我使用NewtonSoft.Json.Net用于序列。

例如，在下面的类我想序列两种特性，但我只想要回地名。

有没有办法做到这一点？

[DataContract]
public class Place
{
   [DataMember(EmitDefaultValue = false)]
   public int PlaceId { get; set; }

   [DataMember(EmitDefaultValue = false, Order = 1)]
   public string PlaceName { get; set; }
}

编辑1：

下面是我当前的JSON文件。

[
  {
    "placeId": 1,
    "placeName": "Malacca"
  },
  {
    "placeId": 2,
    "placeName": "Kuala Lumpur"
  },
  {
    "placeId": 3,
    "placeName": "Genting Highlands"
  },
  {
    "placeId": 4,
    "placeName": "Singapore"
  },
  {
    "placeId": 5,
    "placeName": "Penang"
  },
  {
    "placeId": 6,
    "placeName": "Perak"
  },
  {
    "placeId": 8,
    "placeName": "Selangor"
  }
]

编辑2：找到的解

我发现有一些研究解决方案。

我创建了一个自定义的合同解析器序列化和反序列化的所有属性，并通过它。

下面是我的代码

public  class AllPropertiesResolver : DefaultContractResolver
{
    protected override JsonProperty CreateProperty(MemberInfo member, MemberSerialization memberSerialization)
    {
        JsonProperty property = base.CreateProperty(member, memberSerialization);
        property.Ignored = false;
        return property;
    }
}

而下面是我把它称为代码。

 JsonConvert.SerializeObject(object, new JsonSerializerSettings() { ContractResolver = new AllPropertiesResolver() });
 JsonConvert.DeserializeObject<T>(stream, new JsonSerializerSettings() { ContractResolver = new AllPropertiesResolver() });

谢谢大家的响应。

Answer 1:

您可以使用[JsonIgnore] 既然你标记你的问题asp.net-web-api我想你使用它。所以下面是一个例子，其中，控制器将返回整个模型，除了与属性JsonIgnore 。通过使用自定义ContractResolver我们序列化器它包括所有的属性（即使他们得到了JsonIgnore ）。而使用默认ContractResolver返回我们的反应时。

但要注意，它会覆盖默认行为。所以，你可能要添加一些其他的检查，不仅仅是设置Ignored = false ;

public class PlaceController : ApiController
{
    [HttpGet]
    public IHttpActionResult Get()
    {
        var json = "[{\"placeId\": 1,\"placeName\": \"Malacca\"},{\"placeId\": 2,\"placeName\": \"Kuala Lumpur\"},{\"placeId\": 3,\"placeName\": \"Genting Highlands\"},{\"placeId\": 4,\"placeName\": \"Singapore\"},{\"placeId\": 5,\"placeName\": \"Penang\"},{\"placeId\": 6,\"placeName\": \"Perak\"},{\"placeId\": 8,\"placeName\": \"Selangor\"}]";

        var settings = new JsonSerializerSettings();
        settings.ContractResolver = new IncludeAllPropertiesContractResolver();

        var places = JsonConvert.DeserializeObject<Place[]>(json, settings);
        return Ok(places);
    }
}


public class IncludeAllPropertiesContractResolver : DefaultContractResolver
{
    protected override IList<JsonProperty> CreateProperties(Type type, MemberSerialization memberSerialization)
    {
        IList<JsonProperty> properties = base.CreateProperties(type, memberSerialization);

        // Or other way to determine...
        foreach (var jsonProperty in properties)
        {
            // Include all properties.
            jsonProperty.Ignored = false;
        }
        return properties;
    }
}

[DataContract]
public class Place
{
    [JsonIgnore]
    [DataMember(EmitDefaultValue = false)]
    public int PlaceId { get; set; }

    [DataMember(EmitDefaultValue = false, Order = 1)]
    public string PlaceName { get; set; }
}

输出：

[
{
"placeName": "Malacca"
},
{
"placeName": "Kuala Lumpur"
},
{
"placeName": "Genting Highlands"
},
{
"placeName": "Singapore"
},
{
"placeName": "Penang"
},
{
"placeName": "Perak"
},
{
"placeName": "Selangor"
}
]

或者，如果你不介意一点点思考。下面我们用一个JsonInclude -attribute，这将覆盖的默认行为JsonIgnore 。

public class JsonIncludeContractResolver : DefaultContractResolver
{
    protected override IList<JsonProperty> CreateProperties(Type type, MemberSerialization memberSerialization)
    {
        IList<JsonProperty> properties = base.CreateProperties(type, memberSerialization);

        var actualProperties = type.GetProperties();

        foreach (var jsonProperty in properties)
        {
            // Check if it got our JsonInclude attribute.
            var property = actualProperties.FirstOrDefault(x => x.Name == jsonProperty.PropertyName);
            if (property != null && property.GetCustomAttribute(typeof(JsonInclude)) != null)
            {
                jsonProperty.Ignored = false;
            }
        }
        return properties;
    }
}

[DataContract]
public class Place
{
    [JsonInclude] // Will override JsonIgnore.
    [JsonIgnore]
    [DataMember(EmitDefaultValue = false)]
    public int PlaceId { get; set; }

    [DataMember(EmitDefaultValue = false, Order = 1)]
    public string PlaceName { get; set; }
}

public class JsonInclude : Attribute
{

}

Answer 2:

一个可能的解决方法是使用匿名类： return new { PlaceName = place.PlaceName }; 。

另一种解决方案是创建你自己的序列为您的类型和使用它的类型。例如，对于自定义序列化，你可以找到在这里

Answer 3:

如果您JSON输出是一个List<Place>你可以试试：

        var json = "[{\"placeId\":1,\"placeName\":\"Malacca\"},{\"placeId\":2,\"placeName\":\"Kuala Lumpur\"},{\"placeId\":3,\"placeName\":\"Genting Highlands\"},{\"placeId\":4,\"placeName\":\"Singapore\"},{\"placeId\":5,\"placeName\":\"Penang\"},{\"placeId\":6,\"placeName\":\"Perak\"},{\"placeId\":8,\"placeName\":\"Selangor\"}]";

        var Places = JsonConvert.DeserializeObject<List<Place>>(json);

        foreach (var place in Places)
        {
            Console.WriteLine(place.PlaceName);
        }

文章来源: Custom Json Serializer to serialize and deserialize all properties by ignoring the class attributes