7

想将我的数据序列化为：

<?xml version="1.0" encoding="ibm850"?>
<Batch Name="Test batch">
   <ExecuteCommand Command="..." />
   <WaitCommand Seconds="5" />
</Batch>

但相反，我得到了这个（注意包装命令元素）

<?xml version="1.0" encoding="ibm850"?>
<Batch Name="Test batch">
  <Commands><!-- I want to get rid of thiw wrapper Commands element and just  -->
    <ExecuteCommand Command="..." />
    <WaitCommand Seconds="5" />
  </Commands>
</Batch>

这是用于生成此的示例代码：

public class BaseCommand //base class
{
    [XmlAttribute]
    public string Result { get; set; }
}

public class ExecuteCommand : BaseCommand
{
    [XmlAttribute]
    public string Command { get; set; }
}

public class WaitCommand : BaseCommand
{
    [XmlAttribute]
    public int Seconds { get; set; }
}

public class Batch
{
    [XmlAttribute]
    public string Name { get; set; }

    private List<BaseCommand> _commands = new List<BaseCommand>();
    [XmlArrayItem(typeof(ExecuteCommand))]
    [XmlArrayItem(typeof(WaitCommand))]
    public List<BaseCommand> Commands
    {
        get
        {
            return _commands;
        }
        set
        {
            _commands = value;
        }
    }

    public static void Main()
    {
        XmlSerializer serializer = new XmlSerializer(typeof(Batch));

        Batch b = new Batch();
        b.Name = "Test batch";
        b.Commands.Add(new ExecuteCommand() { Command = "..." });
        b.Commands.Add(new WaitCommand() { Seconds = 5 });

        serializer.Serialize(Console.Out, b);
        Console.Read();
    }
}

我搜索并阅读了有关该主题的大量文章。它们似乎都为使用单一类类型（不使用继承）序列化集合提供了解决方案。我使用继承，似乎没有任何效果。不幸的是，由于遗留支持，我必须输出精确的 XML 文档

4

1 回答 1

9

这是很久以前的事了，但我最终自己想通了。

解决方案是将每个支持的派生类型的 [XmlElement] 属性添加到集合属性 

private List<BaseCommand> _commands = new List<BaseCommand>();
[XmlElement(typeof(ExecuteCommand))]
[XmlElement(typeof(WaitCommand))]
public List<BaseCommand> Commands
{
    get
    {
        return _commands;
    }
    set
    {
        _commands = value;
    }
}
于 2012-11-08T01:10:02.573 回答