Json序列化对象的部分属性值
Json序列化对象的部分属性值在用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？下面介绍一种简单的方法
一、效果如图
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
二、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; }
}
/// <summary>
///简单实体 可变属性序列化器
/// </summary>
public class PropertyVariableJsonSerializer
{
readonly System.Web.Script.Serialization.JavaScriptSerializer _serializer = new JavaScriptSerializer();
/// <summary>
/// json 序列化
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="obj"></param>
/// <param name="propertys"></param>
/// <returns></returns>
public string Serialize<T>(T obj, List<string> propertys)
{
_serializer.RegisterConverters(new[] { new PropertyVariableConveter(typeof(T), propertys) });
return _serializer.Serialize(obj);
}
}
public class PropertyVariableConveter : JavaScriptConverter
{
private readonly List<Type> _supportedTypes = new List<Type>();
public PropertyVariableConveter(Type supportedType, List<string> propertys)
{
_supportedTypes.Add(supportedType);
Propertys = propertys;
}
private List<string> Propertys { get; set; }
public override object Deserialize(IDictionary<string, object> dictionary, Type type, JavaScriptSerializer serializer)
{
throw new Exception(" 这个暂时不支持 ， 谢谢 ");
}
public override IDictionary<string, object> Serialize(object obj, JavaScriptSerializer serializer)
{
var dic = new Dictionary<string, object>();
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<Type> SupportedTypes
{
get { return _supportedTypes; }
}
}
三、调用
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<Person>(p, new List<string>() { "Name", "Age" });
}