Question

我需要将此类转换为XML文件，但由于某些原因，我无法使用.NET序列化库。有没有办法在没有这个工具的情况下转换这个类？

public class Product
{
    public int Id { get; set; }
    public string Title { get; set; }
    public string Desc { get; set; }
    public float Price { get; set; }
    public Category Category { get; set; }
}

我知道XmlSerializer已经尝试过并获得了结果，但我需要在没有XmlSerializer的情况下执行此操作：

Product product = new Product();
var stringwriter = new System.IO.StringWriter();
var serializer = new XmlSerializer(product.GetType());
serializer.Serialize(stringwriter, product);
return stringwriter.ToString();

Answer 1

您只需要使用StringBuilder并连接字符串：

StringBuilder builder = new StringBuilder();

然后：

builder.AppendFormat("<instance property=\"{0}\" />", instance.Property);

如果您需要支持循环参考场景，其中：Product =＆gt; Category =＆gt;产品，然后您需要使用可以为每个实例创建唯一运行时标识的机制，因此您不会多次序列化同一个对象（请查看ObjectIDGenerator）。

希望它有所帮助！

Answer 2

您可以根据自己的要求尝试抽象课程，例如

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            Product p = new Product();
            var x = p.ToXml();
            Console.WriteLine(x.ToString());
            Console.ReadLine();
        }
    }

    public abstract class XmlSerializable
    {
        public XElement ToXml() {
            XElement elm = new XElement(this.GetType().Name);
            this.GetType().GetProperties().ToList().ForEach(p => elm.Add(new XElement(p.Name, p.GetValue(this))));

            return elm;
        }
    }

    public class Product :XmlSerializable
    {
        public int Id { get; set; }
        public string Title { get; set; }
        public string Desc { get; set; }
        public float Price { get; set; }
    }
}

Answer 3

尝试这样的代码

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;

namespace ConsoleApplication2
{
    class Program
    {
         static void Main(string[] args)
         {

             Product product = new Product()
             {
                 id = 123,
                 title = "A Long Way Home",
                 desc = "Welcome",
                 price = 456.78F,
                 category = new Category() { name = "John" }
             };

             XElement xProduct = new XElement("Product", new object[] {
                 new XElement("Id", product.id),
                 new XElement("Title", product.title),
                 new XElement("Description", product.desc),
                 new XElement("Price", product.price),
                 new XElement("Category", product.category)
             });
        }

    }
    public class Product
    {
        public int id { get; set; }
        public string title { get; set; }
        public string desc { get; set; }
        public float price { get; set; }
        public Category category { get; set; }
    }
    public class Category
    {
        public string name { get; set; }
    }
 }

如何在没有序列化工具的情况下将对象转换为XML

3 个答案: