我可以分离JsonConvert对象吗

Question

下面的代码只有FIRST和LAST名称。DeserializeObject正在查找FIRST MIDDLE LAST。所以约翰·史密斯会出现，但约翰·P·史密斯不会。有没有办法只查找名字和姓氏？

#region Fields
private string firstName;
private string middleInitial;
private string lastName;
#endregion
#region Properties
public string FirstName
{
    get { return firstName; }
    set { firstName = value; }
}
public string MiddleInitial
{
    get { return middleInitial; }
    set { middleInitial = value; }
}
public string LastName
{
    get { return lastName; }
    set { lastName = value; }
}
#endregion
#region Constructors
public NamedIndividual()
{
}
var namedIndividual = JsonConvert.DeserializeObject<NamedIndividual>(cov.GetAttribute("individual").Value);

Answer 1

JSON.NET有一种很好的方法，可以将属性标记为非必需属性-您可以使用DataContract和DataMember属性。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Runtime.Serialization;
using Newtonsoft.Json;
namespace ConsoleApplication1
{
    [DataContract]
    public class NamedIndividual
    {
        #region Fields
        private string firstName;
        private string middleInitial;
        private string lastName;
        #endregion
        #region Properties
        [DataMember(IsRequired = true)]
        public string FirstName
        {
            get { return firstName; }
            set { firstName = value; }
        }
        [DataMember(IsRequired = false)]
        public string MiddleInitial
        {
            get { return middleInitial; }
            set { middleInitial = value; }
        }
        [DataMember(IsRequired = true)]
        public string LastName
        {
            get { return lastName; }
            set { lastName = value; }
        }
        #endregion
        public NamedIndividual()
        {
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            string name = "{'FirstName':'David', 'MiddleInitial':'Q', 'LastName':'Hoerster'}";
            string name1 = "{'FirstName':'David', 'LastName':'Hoerster'}";
            var obj = JsonConvert.DeserializeObject<NamedIndividual>(name);
            var obj1 = JsonConvert.DeserializeObject<NamedIndividual>(name1);
            Console.WriteLine(obj.MiddleInitial);
            Console.WriteLine("{0} {1} {2}",obj1.FirstName, obj1.MiddleInitial, obj1.LastName);
        }
    }
}

因此，我将MiddleInitial标记为可选属性，这样我的JSON字符串就可以包含它或不包含它。如果我将属性标记为必需（IsRequired=true），那么我的JSON字符串最好具有该属性，否则将引发异常。

UPDATE我可能暗示您需要使用DataContract属性，以便使某些属性成为不需要或不需要的属性。JSON.net也有自己的一组属性，加上@L.B.下面的评论，不使用任何属性提供了默认行为，这是不需要的。下面是一个使用JsonObject属性的快速示例：

[JsonObject(MemberSerialization=MemberSerialization.OptIn)]
public class NamedIndividual1
{
    [JsonProperty(Required = Required.Always)]
    public string FirstName { get; set; }
    [JsonProperty(Required = Required.Default)]
    public string MiddleInitial { get; set; }
    [JsonProperty(Required = Required.Always)]
    public string LastName { get; set; }
}
class Program
{
    static void Main(string[] args)
    {
        string name2 = "{'FirstName':'David', 'LastName':'Hoerster'}";
        var obj2 = JsonConvert.DeserializeObject<NamedIndividual>(name2);
        Console.WriteLine("{0} {1} {2}", obj2.FirstName, obj2.MiddleInitial, obj2.LastName);
    }
}

（我也不能使用任何属性使所有内容都是可选的。）

希望这能有所帮助！

Answer 2

您可以将构造函数更改为空构造函数或将其kepp为空构造函数，并创建和重载构造函数

public NamedIndividual() 
{
  //empty Constructor
} 
//overloaded constructor
public NamedIndividual(string first, string last)
{
  this.firstname = first;
  this.lastname = last;
} 
 // or pass in the middle as an option