Question

我上课了。它有几个属性可以说10.在这10个中，3个填充了剩下的7个数据是空白的.i.e。空字符串“”使用此link作为参考。我想只显示NON-NULL和NON-EMPTY字符串属性。但最终输出有10个属性。我只希望看到3。

namespace Mynamespace.ValueObjects
{
[DataContract]
public class User
{
      [DataMember(Name ="userID", IsRequired = false,EmitDefaultValue = false)]
    public string userID { get; set; }
      [DataMember(Name ="ssn", IsRequired = false,EmitDefaultValue = false)]
    public string ssn { get; set; }
      [DataMember(Name ="empID", IsRequired = false,EmitDefaultValue = false)]
    public string empID { get; set; }
      [DataMember(Name ="schemaAgencyName", IsRequired = false,EmitDefaultValue = false)]
    public string schemaAgencyName { get; set; }
      [DataMember(Name ="givenName", IsRequired = false,EmitDefaultValue = false)]
    public string givenName { get; set; }
      [DataMember(Name ="familyName", IsRequired = false,EmitDefaultValue = false)]
    public string familyName { get; set; }
      [DataMember(Name ="password", IsRequired = false,EmitDefaultValue = false)]
    public string password { get; set; }
      ....

}

}

我也试过

 [JsonProperty(NullValueHandling = NullValueHandling.Ignore)]

也是属性。没运气。我也喜欢这个

 var t = JsonConvert.SerializeObject(usr, Newtonsoft.Json.Formatting.None,
                                                new JsonSerializerSettings
                                                    {NullValueHandling = NullValueHandling.Ignore});

其中'usr'是User实例。我的意思是没有运气，'t'带回了所有10个属性

{"userID":"vick187","ssn":"","empID":"","schemaAgencyName":"","givenName":"","familyName":"","password":"pwd1234",...}

因此，您只能看到填充了用户ID和密码。但我有ssn，empID等仍然出现。我只想要用户ID和密码。任何帮助将不胜感激。

Answer 1

仅仅装饰属性[JsonProperty(NullValueHandling = NullValueHandling.Ignore)]只应该做你想要的。除非属性设置为空字符串。

只是想知道，为什么需要DataMemeber属性？

以下是工作dotnetfiddle

的链接

using System;
using Newtonsoft.Json;
using System.ComponentModel;

public class Program
{

    public static void Main()
    {
        var user = new User();

        user.UserID = "1234";
        user.ssn = "";

        var settings = new JsonSerializerSettings();

        settings.NullValueHandling = NullValueHandling.Ignore;
        settings.DefaultValueHandling = DefaultValueHandling.Ignore;


        Console.WriteLine(JsonConvert.SerializeObject(user, settings));
    }
}

public class User
{
    [DefaultValue("")]
    public string UserID { get; set; }

    [DefaultValue("")]
    public string ssn { get; set; }

    [DefaultValue("")]
    public string empID { get; set; }

    [DefaultValue("")]
    public string schemaAgencyName { get; set; }

    [DefaultValue("")]
    public string givenName { get; set; }

    [DefaultValue("")]
    public string familyName { get; set; }

    [DefaultValue("")]
    public string password { get; set; }
}

Answer 2

您还可以使用以下两个注释：

[DefaultValue("")]
[JsonProperty(DefaultValueHandling = DefaultValueHandling.Ignore)]
public string Category { get; set; }

Answer 3

我已经使用转换器完成了此操作。

using System;

using Newtonsoft.Json;
using Newtonsoft.Json.Linq;

namespace TestApp.JsonConverterResolver
{
    public class IgnoreEmptyStringsConverter : JsonConverter
    {
    #region Methods

        public override bool CanConvert(Type objectType)
        {
            return objectType == typeof(string);
        }

        public override object ReadJson(JsonReader reader, Type objectType, object existingValue,
                                        JsonSerializer serializer)
        {
            var theValue = reader.Value?.ToString();

            return !string.IsNullOrWhiteSpace(theValue) ? theValue : null;
        }

        public override void WriteJson(JsonWriter writer, object value, JsonSerializer serializer)
        {
            if (!string.IsNullOrWhiteSpace(value.ToString()))
            {
                JToken token = JToken.FromObject(value.ToString(), serializer);
                token.WriteTo(writer);
                return;
            }

            writer.WriteNull();
        }

    #endregion
    }
}

人物模型类示例：

public class Person
{
    public string Name { get; set; }
}

还有语言：

var serializerSettings = new JsonSerializerSettings
     {
         Formatting           = Newtonsoft.Json.Formatting.Indented,
         NullValueHandling    = Newtonsoft.Json.NullValueHandling.Ignore,
         Converters           = new List<JsonConverter> {new IgnoreEmptyStringsConverter()}
     };

var person = JsonConvert.DeserializeObject<Person>("{ \"Name\":\"\" }", serializerSettings);

var jsonPerson = JsonConvert.SerializeObject(new Person { Name = "" }, serializerSettings);

我只是从脑海里写下来。但是我认为这就是我解决问题的方式。也许对某人有帮助。

Answer 4

尽管可接受的答案有效，但它也会删除零值的整数属性。我正在寻找更通用的东西来处理大型对象。

在这里找到了一个很好的答案：https://codearticles.ru/articles/2905?AspxAutoDetectCookieSupport=1

并将其合并为以下用例：

public class ShouldSerializeContractResolver : DefaultContractResolver
{
    public static readonly ShouldSerializeContractResolver Instance = new ShouldSerializeContractResolver();

    protected override JsonProperty CreateProperty(MemberInfo member, MemberSerialization memberSerialization)
    {
        JsonProperty property = base.CreateProperty(member, memberSerialization);

        if (property.PropertyType == typeof(string))
        {
            // Do not include emptry strings
            property.ShouldSerialize = instance =>
            {
                return !string.IsNullOrWhiteSpace(instance.GetType().GetProperty(member.Name).GetValue(instance, null) as string);
            };
        }
        else if (property.PropertyType == typeof(DateTime))
        {
            // Do not include zero DateTime
            property.ShouldSerialize = instance =>
            {
                return Convert.ToDateTime(instance.GetType().GetProperty(member.Name).GetValue(instance, null)) != default(DateTime);
            };
        }
        else if (typeof(IEnumerable).IsAssignableFrom(property.PropertyType))
        {
            // Do not include zero-length lists
            switch (member.MemberType)
            {
                case MemberTypes.Property:
                    property.ShouldSerialize = instance =>
                    {
                        var enumerable = instance.GetType().GetProperty(member.Name).GetValue(instance, null) as IEnumerable;
                        return enumerable != null ? enumerable.GetEnumerator().MoveNext() : false;
                    };
                    break;
                case MemberTypes.Field:
                    property.ShouldSerialize = instance =>
                    {
                        var enumerable = instance.GetType().GetField(member.Name).GetValue(instance) as IEnumerable;
                        return enumerable != null ? enumerable.GetEnumerator().MoveNext() : false;
                    };
                    break;
            }
        }
        return property;
    }
}

可以如下使用：

JsonConvert.SerializeObject(o,
    Newtonsoft.Json.Formatting.None,
    new JsonSerializerSettings
    {
        NullValueHandling = NullValueHandling.Ignore,
        ContractResolver = ShouldSerializeContractResolver.Instance
    });

从json序列化对象

4 个答案: