Question

我无法使用XmlSerializer生成以下XML结构：

<Root xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <Name>This is root.</Name>
  <OtherValue>Otha.</OtherValue>
  <Shapes Name="This attribute is ignored!">
    <Circle>
      <Name>This</Name>
      <Value>Is</Value>
      <Whatever>Circle</Whatever>
    </Circle>
    <Square>
      <Name>And</Name>
      <Value>this is</Value>
      <Something>Square</Something>
    </Square>
  </Shapes>
</Root>

唯一的问题是 <Shapes> 的属性没有被写入。我用于序列化的类如下：

public class Root
{
    [XmlElement]
    public string Name { get; set; }

    [XmlElement]
    public string OtherValue { get; set; }

    [XmlArray("Shapes")]
    [XmlArrayItem("Circle", typeof(Circle))]
    [XmlArrayItem("Square", typeof(Square))]
    public ShapeList Shapes { get; set; }
}

public class ShapeList : List<Shape>
{
    // Attribute that is not in output
    [XmlAttribute]
    public string Name { get; set; }
}

public class Shape
{
    public string Name { get; set; }
    public string Value { get; set; }
}

public class Circle : Shape
{
   public string Whatever { get; set; }
}

public class Square : Shape
{
    public string Something { get; set; }
}

运行序列化的一个主要方法：

public static void Main(String[] args)
{
    var extraTypes = new Type[] {
        typeof(Shape),
        typeof(Square),
        typeof(Circle)
    };

    var root = new Root();
    root.Name = "This is root.";
    root.OtherValue = "Otha.";

    root.Shapes = new ShapeList()
    {
        new Circle() { Name = "This", Value="Is", Whatever="Circle" },
        new Square() { Name = "And", Value="this is", Something="Square" }
    };
    root.Shapes.Name = "This is shapes.";

    using (var sw = new StreamWriter("data.xml"))
    {
        var serializer = new XmlSerializer(typeof(Root), extraTypes);
        serializer.Serialize(sw, root);
    }
}

为什么我没有获取 ShapeList 的 Name 属性？
如果使用此方法无法实现，还有其他简单方法吗？

Answer 1

不处理数组外部的属性;只处理叶子节点 - 集合只是：它们的内容。有一种方法可以做到，如果你不介意让模型更复杂......

public class Root
{
    [XmlElement]
    public string Name { get; set; }

    [XmlElement]
    public string OtherValue { get; set; }

    [XmlElement("Shapes")]
    public ShapeContainer Shapes { get; set; }
}

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

    private readonly List<Shape> items = new List<Shape>();
    [XmlElement("Circle", typeof(Circle))]
    [XmlElement("Square", typeof(Square))]
    public List<Shape> Items { get { return items; } }
}

public class Shape
{
    public string Name { get; set; }
    public string Value { get; set; }
}

public class Circle : Shape
{
    public string Whatever { get; set; }
}

public class Square : Shape
{
    public string Something { get; set; }
}

使用情况更改如下：

root.Shapes = new ShapeContainer();
root.Shapes.Items.Add(new Circle() { Name = "This", Value="Is", Whatever="Circle" });
root.Shapes.Items.Add(new Square() { Name = "And", Value = "this is", Something = "Square" });
root.Shapes.Name = "This is shapes.";

XmlAttribute不能与XmlArray一起使用

1 个答案: