Question

XML

<MeterWalkOrder>
  <Name>Red Route</Name>
  <Meters>
    <Meter>
      <MeterID>1</MeterID>
      <SerialNumber>12345</SerialNumber>
    </Meter>
    <Meter>
      <MeterID>2</MeterID>
      <SerialNumber>SE</SerialNumber>
    </Meter>
  </Meters>
</MeterWalkOrder>

我无法使用任何序列化程序将简单的XML转换为对象

var xml = File.ReadAllText("WalkOrder.xml");
var xmlSerializer = new NFormats.Xml.XmlSerializer();
var obj = xmlSerializer.Deserialize<MeterWalkOrder>(new StringReader(xml));

我刚刚找回没有设置属性的2米对象，甚至没有按步行顺序设置名称。

 public partial class MeterWalkOrder
 {
        public MeterWalkOrder()
        {
            Meters = new List<Meter>();
        }

        [DataMember]
        public String Name { get; set; }
     }
}

using System;
using System.Xml.Serialization;

namespace WindowsFormsApplication1.Classes
{    
    public class Meter : IMeter
    {
        [XmlAttribute]
        public int MeterID { get; set; }

        [XmlAttribute]
        public String SerialNumber { get; set; }
    }
}

我愿意尝试任何xml序列化程序。

Answer 1

首先，我建议你阅读Introducing XML Serialization on MSDN 您犯了一些错误，导致在运行代码时抛出未提及的异常。

不是它们是元素的属性。（如Wyat Earp评论）
如果你想序列化某些东西，你必须告诉它应该是[Serializable]
序列化需要实现的公共空构造函数
当你没有关闭它们时不要打开流（使用“使用”）
要测试序列化是否最佳，首先序列化，然后检查输出然后实现反序列化

在下面找到一个完整的例子：

using System;
using System.Collections.Generic;
using System.IO;
using System.Xml;
using System.Xml.Serialization;

namespace X123
{
    static class Program
    {
        /// <summary>
        /// The main entry point for the application.
        /// </summary>
        [STAThread]
        static void Main()
        {
            MeterWalkOrder mo = new MeterWalkOrder();
            mo.Name = "name";
            mo.Meters.Add(new Meter { MeterID = 1, SerialNumber = "kdkdkd" });
            mo.Meters.Add(new Meter { MeterID = 2, SerialNumber = "holladrio" });

            var xmlSerializer = new XmlSerializer(typeof(MeterWalkOrder), new Type[] { typeof(Meter) });
            {
                xmlSerializer.Serialize(File.CreateText("hello.xml"), mo);
                using (Stream s = File.OpenRead("hello.xml"))
                {
                    var obj = xmlSerializer.Deserialize(s);
                }
            }
        }



    }

    [Serializable]
    public class MeterWalkOrder
    {
        public MeterWalkOrder()
        {
        }

        public string Name { get; set; }
        public List<Meter> Meters { get { return meters; } set { meters = value; } }
        private List<Meter> meters = new List<Meter>();
    }

    [Serializable]
    public class Meter
    {
        public Meter()
        {
        }

        [XmlAttribute]
        public int MeterID { get; set; }

        [XmlAttribute]
        public string SerialNumber { get; set; }
    }

Answer 2

我使用了您的示例XML并使用Paste Special -> Paste XML as Classes在VisualStudio中生成了类，对它们进行了一些修改以使它们更具可读性并从中获得以下类定义：

[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "", IsNullable = false)]
public class MeterWalkOrder
{
  public string Name { get; set; }

  [System.Xml.Serialization.XmlArrayItemAttribute("Meter", IsNullable = false)]
  public List<MeterWalkOrderMeter> Meters { get; set; }
}

[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
public class MeterWalkOrderMeter
{
  public int MeterID { get; set; }
  public string SerialNumber { get; set; }
}

使用上面的类和下面的代码，它生成了没有错误的对象。

string inputXml = File.ReadAllText(@"C:\Temp\SOTest.xml");
//using System.Xml.Serialization;
System.Xml.Serialization.XmlSerializer xmlSerializer = new System.Xml.Serialization.XmlSerializer(typeof(MeterWalkOrder));
MeterWalkOrder outputObject = xmlSerializer.Deserialize(new StringReader(inputXml)) as MeterWalkOrder;

无法将简单的XML解析为对象？

2 个答案: