C# XML Serializing list with element name depending on class

Question

I want to serialize a List<AbstractClass> that contains DerivedClassA and DerivedClassB where both of these classes are derived from AbstractClass.

Additionally I want to add an attribute to the element that wraps each element of the list. My desired output is something like this:

<MyCustomListName foo="bar">
    <DerivedClassA baseAttribute="value" attributeFromA="value"/>
    <DerivedClassB baseAttribute="value" attributeFromB="value"/>
    ...
</MyCustomListName>

What I have so far is this:

public class MyCustomListName
{
   [XmlAttribute]
   public string foo {get; set;}

   [XmlArray]        
   public List<AbstractClass> list { get; set; }

}

[XmlInclude(typeof(DerivedClassA))]
[XmlInclude(typeof(DerivedClassB))]
public abstract class AbstractClass
{
   [XmlAttribute]
   public string baseAttribute {get; set;}
}

public class DerivedClassA : AbstractClass
{
   [XmlAttribute]
   public string attributeFromA {get; set;}
}

public class DerivedClassB : AbstractClass
{
   [XmlAttribute]
   public string attributeFromB {get; set;}
}

Using this, I get the following output:

<MyCustomListName foo="bar">
  <list>
    <AbstractClass xsi:type="DerivedClassA" baseAttribute="value" attributeFromA="value"/>
    <AbstractClass xsi:type="DerivedClassB" baseAttribute="value" attributeFromB="value"/>
  </list>
</MyCustomListName>

So, to get my desired output, there should be no nested <list> element and instead of the xsi:type attribute, the element name should be DerivedClassA or DerivedClassB depending on the type.

Sergey Berezovskiy · Accepted Answer · 2017-01-16 13:04:57Z

3

Provide different names for items of different types:

[XmlElement("derivedA", typeof(DerivedClassA))]
[XmlElement("derivedB", typeof(DerivedClassB))]
public List<AbstractClass> list { get; set; }

edited Jan 16, 2017 at 13:04

answered Jan 16, 2017 at 12:53

Sergey Berezovskiy

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1 Comment

Eddi Over a year ago

This seems to work for creating the proper element names. However, I would need to not output the <list> tag.

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