XmlSerializer - Array in array

Question

I have the following xml file:

<ArrayOfX xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <X>
    <Name>Name1</Name>
    <ArrayOfY>
      <Y>
        <Member1>1</Member1>
      </Y>
      ...
    </ArrayOfY>
  </X>
  ...
</ArrayOfX>

public class X {
    public string Name { get; set; }
    public List<Y> Y { get; set; }
}

public class Y {
    public String Member1 { get; set; }
}

But if i try to deserialize with XmlSerializer, Name contains the correct value but the Y list is empty. Any idea?

XmlSerializer serializer = new XmlSerializer( typeof( List<X> ) );
return (List<X>)serializer.Deserialize(reader);

Fung · Accepted Answer · 2012-10-24 11:19:20Z

1

Try this:

public class X {
    public string Name { get; set; }
    [XmlArray(ElementName = "ArrayOfY")]
    public List<Y> Y { get; set; }
}

With you class definition, if you try to serialize of a List<X>, the output would looks like:

<ArrayOfX xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
  <X>
    <Name>Name1</Name>
    <Y> <!-- Not ArrayOfY -->
      <Y>
        <Member1>1</Member1>
      </Y>
      ...
    </Y>
  </X>
  ...
</ArrayOfX>

Your provided xml file is not a valid format to the XmlSerializer(typeof(List<X>)). It's expecting <Y> instead of <ArrayOfY> for the List<Y> member of X.

The attribute I added would instruct the serializer to look for an element with name ArrayOfY instead.

edited Oct 24, 2012 at 11:19

answered Oct 24, 2012 at 10:57

Fung

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Comments

SpruceMoose · Accepted Answer · 2012-10-24 10:34:39Z

1

You will need to populate the 'extraTypes' array parameter with typeof(Y).

See http://msdn.microsoft.com/en-us/library/e5aakyae.aspx

answered Oct 24, 2012 at 10:34

SpruceMoose

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1 Comment

Aaaaaaaa Over a year ago

I've already tried it: XmlSerializer serializer = new XmlSerializer( typeof( List<X> ), new Type[] { typeof( Y ), typeof( List<Y> ) } );, but does not work

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