在用JSON序列化对象是，会返回这个对象的所有属性键值对，如果某个对象的属性非常多，但是我们需要获取的JSON数据只是其中的两三个属性，这样的情况，我们怎么优化呢？

具体实现方式

using System;    using System.Collections.Generic;    using System.Web.Script.Serialization;    public class Person    {        public string Name { get; set; }        public int Age { get; set; }        public double Meney { get; set; }        public double Tex { get; set; }        public DateTime Berthday { get; set; }    }    /** 
    ///简单实体 可变属性序列化器    /// 
    public class PropertyVariableJsonSerializer    {        readonly System.Web.Script.Serialization.JavaScriptSerializer _serializer = new JavaScriptSerializer();        /** 
        /// json 序列化        /// 
        /// 
            ///         ///         /// 
            public string Serialize
    
     (T obj, List
     
       propertys)        {            _serializer.RegisterConverters(new[] { new PropertyVariableConveter(typeof(T), propertys) });            return _serializer.Serialize(obj);        }    }    public class PropertyVariableConveter : JavaScriptConverter    {        private readonly List
      
        _supportedTypes = new List
       
        ();        public PropertyVariableConveter(Type supportedType, List
        
          propertys)        {            _supportedTypes.Add(supportedType);            Propertys = propertys;        }        private List
         
           Propertys { get; set; } public override object Deserialize(IDictionary
          
            dictionary, Type type, JavaScriptSerializer serializer) { throw new Exception(" 这个暂时不支持 ， 谢谢 "); } public override IDictionary
           
             Serialize(object obj, JavaScriptSerializer serializer) { var dic = new Dictionary
            
             (); var t = obj.GetType(); var properties = t.GetProperties(); foreach (var ite in properties) { string key = ite.Name; var v = t.GetProperty(key).GetValue(obj, null); if (Propertys == null || Propertys.Count <= 0) { dic.Add(key, v); continue; } if (Propertys.Contains(key)) { dic.Add(key, v); } } return dic; } public override IEnumerable
             
               SupportedTypes { get { return _supportedTypes; } } }
             
            
           
          
         
        
       
      
     
    

调用

在序列化一个对象时， 只序列化了我们想要的两个属性，  实际对象有4个属性

public static void aaa()        {            var p = new Person { Age = 20, Name = "www.studyofnet.com", Meney = 3, Tex = 1};            var s = new PropertyVariableJsonSerializer();            string result = s.Serialize
    
     (p, new List
     
      () { "Name", "Age" });        }
     
    

参考资料：如何Json序列化对象的部分属性