序列化不带类名的 XML 列表元素
本文关键字:XML 列表元素 序列化 | 更新日期: 2023-09-27 17:56:26
I 以下类:
public class Family
{
public List<ChildAge> childAges { get; set; }
}
现在,ChildAge 看起来像这样:
public class ChildAge
{
public int Age { get; set; }
}
当我将其序列化为 XML 时,我得到:
<root>
<Family>
<ChildAges>
<ChildAge>
<Age>10</Age>
</ChildAge>
<ChildAge>
<Age>8</Age>
</ChildAge>
</ChildAges>
</Family>
</root>
我需要更改什么才能获得此内容:
<root>
<Family>
<ChildAges>
<Age>10</Age>
<Age>8</Age>
</ChildAges>
</Family>
<root>
谢谢!
一旦您在"childage"下有更多属性要序列化,您的问题就会开始您可以简单地使用列表和以下注释:
[XmlArray("ChildAges")]
[XmlArrayItem("Age")]
List<int> ChildrenAges { get; set; }
你差不多完成了
您可以通过提供自己的 WriteXML 实现来做到这一点(以及更多)。
请看下面的代码段。
public class Family : IXmlSerializable
{
public List<ChildAge> childAges { get; set; }
public void WriteXml(XmlWriter writer)
{
foreach(ChildAge ca in childAges)
writer.WriteElementString("Age", ca.Age.ToString());
}
public void ReadXml(XmlReader reader)
{
// [...]
}
public XmlSchema GetSchema()
{
return (null);
}
}
public class ChildAge
{
public int Age { get; set; }
}
public class Program
{
static void Main(string[] args)
{
Family f = new Family();
f.childAges = new List<ChildAge>();
f.childAges.Add(new ChildAge() { Age = 10 });
f.childAges.Add(new ChildAge() { Age = 8 });
XmlSerializer xs = new XmlSerializer(typeof(Family));
XmlSerializerNamespaces xmlnsEmpty;
xmlnsEmpty = new XmlSerializerNamespaces(new[] { XmlQualifiedName.Empty });
XmlWriterSettings writerSettings = new XmlWriterSettings();
writerSettings.Indent = true;
writerSettings.OmitXmlDeclaration = true;
StringBuilder sb = new StringBuilder();
XmlWriter writer = XmlTextWriter.Create(sb, writerSettings);
xs.Serialize(writer, f, xmlnsEmpty);
Console.WriteLine(sb.ToString());
Console.ReadLine();
}
}