c# - C# XML 序列化每个元素的属性

Question

我想创建以下 xml：

 <StartLot>
      <fileCreationDate level="7">201301132210</fileCreationDate>
      <fmtVersion level="7">3.0</fmtVersion>
 </StartLot>

下面是序列化代码：

[Serializable]
class StartLot
{
    public fileCreationDate{get; set;}

    [XmlAttribute("level")] 
    public string level = "7";

    public fmtVersion{get; set;}

    [XmlAttribute("level")] 
    public string level = "7"; ??
}

由于我已经声明了属性级别，如何添加最后一个属性？

Answer 1

您可以使用包装类来存储这两个值，如下例所示：

public class StackOverflow_15441384
{
    const string XML = @"<StartLot>
                               <fileCreationDate level=""7"">201301132210</fileCreationDate>
                               <fmtVersion level=""7"">3.0</fmtVersion>
                            </StartLot>";
    public class StartLot
    {
        [XmlElement("fileCreationDate")]
        public LevelAndValue FileCreationDate { get; set; }
        [XmlElement("fmtVersion")]
        public LevelAndValue FmtVersion { get; set; }
    }
    public class LevelAndValue
    {
        [XmlAttribute("level")]
        public string Level { get; set; }
        [XmlText]
        public string Value { get; set; }
    }
    public static void Test()
    {
        XmlSerializer xs = new XmlSerializer(typeof(StartLot));
        StartLot sl = (StartLot)xs.Deserialize(new MemoryStream(Encoding.UTF8.GetBytes(XML)));
        Console.WriteLine("FCD.L = {0}", sl.FileCreationDate.Level);
        Console.WriteLine("FCD.V = {0}", sl.FileCreationDate.Value);
        Console.WriteLine("FV.L = {0}", sl.FmtVersion.Level);
        Console.WriteLine("FV.V = {0}", sl.FmtVersion.Value);
    }
}

Answer 2

我总是觉得Linq To Xml更容易使用

var xDoc = XDocument.Parse(xml); /* XDocument.Load(filename); */

var items = xDoc.Root.Descendants()
                .Select(e => new
                {
                    Name = e.Name.LocalName,
                    Level = e.Attribute("level").Value,
                    Value = e.Value
                })
                .ToList();