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简单的XML和JSON数据的处理

时间:2016-06-05 16:58:10      阅读:191      评论:0      收藏:0      [点我收藏+]

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一、XML格式装换成json数据格式

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
//要使用JavaScriptSerializer需添加这个using System.Web.Script.Serialization;引用
using System.Web.Script.Serialization;
using System.Web.UI;
using System.Web.UI.WebControls;
//要使用XDocument需添加这个using System.Xml.Linq引用
using System.Xml.Linq;
//要使用JsonConvert需添加这个Newtonsoft.Json.dll引用
using Newtonsoft.Json;
using System.Xml;

public partial class xmlchange : System.Web.UI.Page
{
    protected void Page_Load(object sender, EventArgs e)
    {
        var xml =
        @"<Columns>
          <Column Name=""key1"">key1</Column>
          <Column Name=""key2"">key2</Column>
          <Column Name=""key3"">key3</Column>
        </Columns>";
        var dic = XDocument
            .Parse(xml)
            .Descendants("Column")
            .ToDictionary(
                c => c.Attribute("Name").Value,
                c => c.Value
            );
        var json = new JavaScriptSerializer().Serialize(dic);
        Label1.Text = json;

        xml model = JsonConvert.DeserializeObject<xml>(json);
        Label2.Text = model.key1;
        Label3.Text = model.key2;
        Label4.Text = model.key3;
        
    }
    public class xml
    {
        public string key1 { get; set; }
        public string key2 { get; set; }
        public string key3 { get; set; }
    }
}

二,json数据格式的序列化和反序列化

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Script.Serialization;
using System.Web.UI;
using System.Web.UI.WebControls;

public partial class jsonchange : System.Web.UI.Page
{
    protected void Page_Load(object sender, EventArgs e)
    {
        test te = new test();
        te.test1 = 1;
        te.test2 = "test";


        JavaScriptSerializer jsonSerializer = new JavaScriptSerializer();
        //执行序列化
        string r1 = jsonSerializer.Serialize(te);
        Label1.Text =r1;
        //执行反序列化
        test per = jsonSerializer.Deserialize<test>(r1);
        Label2.Text = per.test1.ToString();
        Label3.Text = per.test2;
    }
    public class test
    {
        public int test1 { get; set; }
        public string test2 { get; set; }
    }
}

 

简单的XML和JSON数据的处理

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原文地址:http://www.cnblogs.com/May-day/p/5561055.html

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