13

Morning Guys,

I have a collection that descends from List and has a public property. The Xml serializer does not pick up my proeprty. The list items serialize fine. I have tried the XmlAttribute attribute to no avail. Do you guys have a solution?

    public partial class MainWindow : Window
{
    public MainWindow()
    {
        InitializeComponent();
    }
    private void button1_Click(object sender, RoutedEventArgs e)
    {
        var people = new PersonCollection
        {
            new Person { FirstName="Sue", Age=17 },
            new Person { FirstName="Joe", Age=21 }
        };
        people.FavoritePerson = "Sue";

        var x = new XmlSerializer(people.GetType());
        var b = new StringBuilder();
        var w = XmlTextWriter.Create(b, new XmlWriterSettings { NewLineChars = "\r\n", Indent = true });
        x.Serialize(w, people);
        var s = b.ToString();
    }
}

[XmlRoot(ElementName="People")]
public class PersonCollection : List<Person>
{
    //DOES NOT WORK! ARGHHH
    [XmlAttribute]
    public string FavoritePerson { get; set; }    
}

public class Person
{
    [XmlAttribute]
    public string FirstName { get; set; }
    [XmlAttribute]
    public int Age { get; set; }
}

I'm getting the following xml

<?xml version="1.0" encoding="utf-16"?>
        <People xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
          <Person FirstName="Sue" Age="17" />
          <Person FirstName="Joe" Age="21" />
        </People>

I would like to get this

<?xml version="1.0" encoding="utf-16"?>
        <People FavoritePerson="Sue" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
          <Person FirstName="Sue" Age="17" />
          <Person FirstName="Joe" Age="21" />
        </People>

3 Answers 3

5

I went ahead and solved the problem by implementing IXmlSerializable. If a simpler solution exists, post it!

    [XmlRoot(ElementName="People")]
public class PersonCollection : List<Person>, IXmlSerializable
{
    //IT WORKS NOW!!! Too bad we have to implement IXmlSerializable
    [XmlAttribute]
    public string FavoritePerson { get; set; }

    public System.Xml.Schema.XmlSchema GetSchema()
    {
        return null;
    }
    public void ReadXml(XmlReader reader)
    {
        FavoritePerson = reader[0];            
        while (reader.Read())
        {
            if (reader.Name == "Person")
            {
                var p = new Person();
                p.FirstName = reader[0];
                p.Age = int.Parse( reader[1] ); 
                Add(p);
            }
        }
    }
    public void WriteXml(XmlWriter writer)
    {
        writer.WriteAttributeString("FavoritePerson", FavoritePerson);
        foreach (var p in this)
        {
            writer.WriteStartElement("Person");
            writer.WriteAttributeString("FirstName", p.FirstName);
            writer.WriteAttributeString("Age", p.Age.ToString());
            writer.WriteEndElement();            
        }
    }
}
Sign up to request clarification or add additional context in comments.

Comments

3

This isn't an answer to the question, but I thought I'd make a suggestion to ease in code development.

Add a new Add method to the PersonCollection class as such:

public class PersonCollection : List<Person>, IXmlSerializable
{
...
    public void Add(string firstName, int age)
    {
        this.Add(new Person(firstName, age));
    }
...
}

Then, by doing this, you can simplify your collection initializer syntax to:

var people = new PersonCollection
{
    { "Sue", 17 },
    { "Joe", 21 }
};
people.FavoritePerson = "Sue";

Comments

1

If you don't mind having to wrap all of the list functions, then you can embed the list as a property of the class rather than deriving from it.

You'd then use the XmlElement attribute to force the xml elements to be written out as a flat list (rather than being nested).

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.