Let’s assume we have the following XML structure:
<?xml version="1.0" encoding="utf-8" ?>
<Root>
<Node>
<Label Id="Label1">Betreff</Label>
<TextBox Id="TextBox1" MultiLines="3" />
</Node>
<Node>
<Label Id="Label2">Betreff 123132</Label>
<TextBox Id="TextBox2" />
<Button Id="Button1" Label="Hello World" Action="Foobar" />
</Node>
</Root>
Under
Code: Deserialize the elements & co.
The first step: We need to deserialize the
public class Root
{
[XmlElement(ElementName = "Node")]
public List<Node> Nodes { get; set; }
}
Next - and this is the main part of this blogpost - we need to describe the
public class Node
{
[XmlElement(typeof(LabelElement), ElementName = "Label")]
[XmlElement(typeof(TextBoxElement), ElementName = "TextBox")]
[XmlElement(typeof(ButtonElement), ElementName = "Button")]
public List<BaseElement> Elements { get; set; }
}
As you can see, we need to “register” each specialized known element. The default XmlSerializer will read all XmlElement Attributes and will do the main work.
The last part: The base class and the specialized classes for each element.
public abstract class BaseElement
{
[XmlAttribute(AttributeName = "Id")]
public string Id { get; set; }
}
public class ButtonElement : BaseElement
{
[XmlAttribute(AttributeName = "Label")]
public string Label { get; set; }
[XmlAttribute(AttributeName = "Action")]
public string Action { get; set; }
}
public class LabelElement : BaseElement
{
[XmlText]
public string Content { get; set; }
}
public class TextBoxElement : BaseElement
{
[XmlAttribute(AttributeName = "MultiLines")]
public int MultiLines { get; set; }
}
Interfaces, base classes etc.
It doesn’t matter if you choose a interface, base class or a abstract base class. The built-in XmlSerializer is flexible, but as far as I know you will need to “register” the implementation elements on the base element. So there is no “convention” or any magic in place.
Using the code
Using the XmlSerializer is simple:
FileStream readFileStream = new FileStream(@"test.xml", FileMode.Open, FileAccess.Read, FileShare.Read);
XmlSerializer serializer = new XmlSerializer(typeof(Root));
var test = serializer.Deserialize(readFileStream);
Serializing
If you want to serialize objects to XML with this code the XmlSerializer should be well prepared and no code changes should be needed. It just works ;)
Hope this helps!
The code is also available on GitHub.
