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如何使用XmlSerializer为从基类继承的类的节点名编写XmlType

xmlserializer xml c#

0

Doug Clutter · 技术社区 · 10 年前

我想从一组C#对象生成以下XML:

<?xml version="1.0" encoding="utf-16"?>
<MyRoot xmlns="http://sample.com">
  <Things>
    <This>
      <ThisRef>this ref</ThisRef>
    </This>
    <That>
      <ThatRef>that ref</ThatRef>
    </That>
  </Things>
</MyRoot>

不幸的是,XML节点东西可以包含几个这或那个节点,并且我对XML模式没有任何控制权。我需要创建能够编写正确XML的C#对象,但我在东西收集

以下是我目前的情况:

[XmlRoot("MyRoot", Namespace = XmlNamespace)]
public class MyRoot
{
    public const string XmlNamespace = "http://sample.com";
    public List<MyThing> Things { get; set; }

}
[XmlInclude(typeof(This))]
[XmlInclude(typeof(That))]
public class MyThing
{
}
[XmlRoot(Namespace = MyRoot.XmlNamespace)]
[XmlType("This")]
public class This : MyThing
{
    public string ThisRef { get; set; }
}
[XmlRoot(Namespace = MyRoot.XmlNamespace)]
[XmlType("That")]
public class That : MyThing
{
    public string ThatRef { get; set; }
}
[TestClass]
public class so
{
    [TestMethod]
    public void SerializeTest()
    {
        // Create object to serialize
        var data = new MyRoot { Things = new List<MyThing>() };
        data.Things.Add(new This { ThisRef = "this ref" });
        data.Things.Add(new That { ThatRef = "that ref" });

        // Create XML namespace
        var xmlNamespaces = new XmlSerializerNamespaces();
        xmlNamespaces.Add(string.Empty, MyRoot.XmlNamespace);

        // Write XML
        var serializer = new XmlSerializer(typeof(MyRoot));
        using (var writer = new StringWriter())
        {
            serializer.Serialize(writer, data, xmlNamespaces);
            var xml = writer.ToString();
            Console.WriteLine(xml);
        }
    }
}

生成以下XML:

<?xml version="1.0" encoding="utf-16"?>
<MyRoot xmlns="http://sample.com">
  <Things>
    <MyThing d3p1:type="This" xmlns:d3p1="http://www.w3.org/2001/XMLSchema-instance">
      <ThisRef>this ref</ThisRef>
    </MyThing>
    <MyThing d3p1:type="That" xmlns:d3p1="http://www.w3.org/2001/XMLSchema-instance">
      <ThatRef>that ref</ThatRef>
    </MyThing>
  </Things>
</MyRoot>

所以,我有两个问题:

如何使XmlSerializer写入这和那个节点,而不是 我的东西 节点?
如何阻止XmlSerializer向这和那个节点(当前写为 我的东西 节点)?

1 回复 | 直到 10 年前

1

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Doug Clutter 10 年前

在stackoverflow上发帖是什么让你立即找到答案?我只需要做以下几点:

[XmlRoot("MyRoot", Namespace = XmlNamespace)]
public class MyRoot
{
    public const string XmlNamespace = "http://sample.com";
    [XmlArrayItem(Type = typeof(This))]
    [XmlArrayItem(Type = typeof(That))]
    public List<MyThing> Things { get; set; }
}
public class MyThing
{
}
public class This : MyThing
{
    public string ThisRef { get; set; }
}
public class That : MyThing
{
    public string ThatRef { get; set; }
}