序列化派生类型 - 不在数组中

Question

通过 Visual Studio 设置设计器（自动生成的代码）使用 .NET XmlSerializer，可以像这样序列化派生类型的数组：

[XmlArrayItem(Type = typeof(Square)), XmlArrayItem(Type = typeof(Triangle))]
public Shape[] Shapes;

...

Properties.Settings.Default.Save();

这会产生 XML 就像

<Shape>
  <Triangle>....</Triangle>
  <Triangle>....</Triangle>
  <Square>....</Square>
</Shape>

但是如果派生类型不在数组（或任何其他集合）中怎么办？

将XmlArrayItem替换为XmlElement，如下

[XmlElement(Type = typeof(Square)), XmlElement(Type = typeof(Triangle))]
public Shape Window;

有效但产生

<Triangle>....</Triangle>

其中元素名称是类型，而不是属性名称，如果我添加第二个形状：

[XmlElement(Type = typeof(Square)), XmlElement(Type = typeof(Triangle))]
public Shape Door;
[XmlElement(Type = typeof(Square)), XmlElement(Type = typeof(Triangle))]
public Shape Window;

序列化只是失败了——毕竟，它如何在反序列化时知道哪个 XML 元素进入哪个属性？

我缺少某个属性吗？我可以在不编写代码来自定义序列化的情况下完成这项工作吗？怎么样？

Answer 1

复合形状序列化初稿：

  [Serializable()]
  [XmlRoot("shape", Namespace = "", IsNullable = false)]
  public abstract class Shape {
    public abstract void Draw();
    [XmlAttribute("name")] public string Name { get; set; }
  }

  [Serializable()]
  [XmlRoot("triangle", Namespace = "", IsNullable = false)]
  public class Triangle : Shape {
    public override void Draw() { }
  }

  [Serializable()]
  [XmlRoot("square", Namespace = "", IsNullable = false)]
  public class Square : Shape {
    public override void Draw() { }
  }

  [Serializable()]
  [XmlRoot("compositeShape", Namespace = "", IsNullable = false)]
  public class CompositeShape : Shape {
    [XmlElement("shape", typeof(Shape))]
    [XmlElement("triangle", typeof(Triangle))]
    [XmlElement("square", typeof(Square))]
    [XmlElement("compositeShape", typeof(CompositeShape))]
    public Shape[] Items { get; set; }

    public override void Draw() { }
  }

像这样使用它：

var shape = new CompositeShape() {
    Name = "some composite shape",
    Items = new Shape[] {
      new CompositeShape() { 
        Name = "inner composite shape",
        Items = new Shape[] {
          new Triangle() {Name = "level 2 triangle"},
          new Square() {Name="level 2 square"}
        } }, 
        new Triangle() {Name = "level 1 triangle"},
        new Square() {Name="level 1 square"}
    }};
    // serialize ...

示例输出：

<compositeShape name="some composite shape">
  <compositeShape name="inner composite shape">
    <triangle name="level 2 triangle" />
    <square name="level 2 square" />
  </compositeShape>
  <triangle name="level 1 triangle" />
  <square name="level 1 square" />
</compositeShape>

这种方法的缺点是，每次将对象添加到层次结构时，您还必须使用该类型的元素来装饰形状数组，因此不会真正关闭以进行修改...

Answer 2

答案是在基础 class 上使用 XmlInclude 而不是在属性上使用 XmlElement：

[XmlInclude(typeof(Triangle))]
[XmlInclude(typeof(Square))]
public abstract class Shape

生成的内容略有不同 XML。对于数组情况：

<Shapes>
  <Shape xsi:type="Triangle">....</Shape>
  <Shape xsi:type="Triangle">....</Shape>
  <Shape xsi:type="Square">....</Shape>
</Shapes>

对于标量情况：

<Door xsi:type="Square">....</Door>
<Window xsi:type="Triangle">....</Window>

完美。

Answer 3

添加 [XmlInclude] attributes on the root object as indicated in 是一个很好的解决方案，因为它可以处理数据模型中任何地方出现的所有 Square 类型的多态属性。

然而，这种情况也可以使用 [XmlElement] attributes by disambiguating the polymorphic elements via XmlElementAttribute.ElementName:

来处理

public class Room
{
    [XmlElement("DoorSquare", Type = typeof(Square)), XmlElement("DoorTriangle", Type = typeof(Triangle))]  
    public Shape Door { get; set; }

    [XmlElement("WindowSquare", Type = typeof(Square)), XmlElement("WindowTriangle", Type = typeof(Triangle))]  
    public Shape Window { get; set; }
}

这会产生以下结果 XML：

<Room>
  <DoorTriangle />
  <WindowSquare />
</Room>

示例 fiddle.

序列化派生类型 - 不在数组中

Serializing derived types - not in an array

c#

xmlserializer