.NET XML Deserialization

Question

I have the following XML feed. I need to do deserialization to a List of objects. How can I achieve this in .NET 4.0 C#?

<body>
  <games>
     <sports>
       <lot_name>Football</lot_name>
       <prizes>
           <!-- For each prize the "division_" tag ends with different number -->
           <divisions_1>
              <divisions>1</divisions>
              <match>5-2</match>
              <pay>$10</pay>
           </divisions_1>
           <divisions_2>
              <divisions>2</divisions>
              <match>3-2</match>
              <pay>$5</pay>
          </divisions_2>
       </prizes>
     </sports>
  </games>
</body>

You won't get a collection; you'll get a single `body` object composed of a single `games` object composed of a single `sports` object composed of a single `prizes` object composed of a `divisions_1` object and a `divisions_2` object... — Brian Driscoll, Sep 02 '11 at 13:13
Is there a variable number of `` tags? If so, can you use XSLT to change them to each be `` instead? You're going to have a hard time deserializing to a class if the tags aren't consistent. — Yuck, Sep 02 '11 at 13:13
@DavidBonnici in your original post you were missing the closing prizes tag. I've since amended my comment. — Brian Driscoll, Sep 02 '11 at 13:16
Your best bet here will be to change the `` through `` to just ``. This will make serializing to a List much easier. This should be ok to do, since you have the `divisions` number as a sub element. — SwDevMan81, Sep 02 '11 at 13:17

score 1 · Accepted Answer · edited May 23 '17 at 11:43

1

Look at here, a good explanation.

xsd.exe is a handy little tool that comes with the .NET SDK (Software Development Kit) that can make life easy when you want to quickly go from an XML input to a strongly-typed .NET object.

If you're starting with an XML file:

1) Open the XML file in Visual Studio

2) From the XML menu click "Create Schema". This will generate a XSD file.

The rest here.

edited May 23 '17 at 11:43

Community

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answered Sep 02 '11 at 13:13

CharithJ

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The link provided to http://blog.stevehorn.cc/2007/07/generate-code-from-xml-schema-xsdexe.html no longer works. – Tangiest Mar 10 '15 at 01:19

Christoph Fink · Answer 2 · 2011-09-02T13:19:25.077

0

Take a look at THIS tutorial on how to do XML serialization, the example there is pretty near to yours.

Basically you need to create a class which represents each entity in your XML (e.g. Sports) and the corresponding properties and then us an XmlSerializer to deserialze it.

The more complex way, but where you can handle the strange <division_X> tags, would be to use an XmlReader and read in the elements on your own and assign it to appropriate classes/objects.

edited Sep 02 '11 at 13:19

answered Sep 02 '11 at 13:12

Christoph Fink

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Not even remotely close, apart from both using XML. The issue is going to be with the variable `` tags. – Yuck Sep 02 '11 at 13:14

score 0 · Answer 3 · answered Sep 02 '11 at 14:47

Unless you want to generate a XSD with N division_x elements where x = [1..N] you have to implement IXmlSerializable. The following code does deserialize the example xml nicely:

public class Sports : IXmlSerializable
{
    public string LotName { get; set; }

    public List<dynamic> Prices { get; set; }

    #region IXmlSerializable Members

    public System.Xml.Schema.XmlSchema GetSchema()
    {
        return null;
    }

    public void ReadXml(System.Xml.XmlReader reader)
    {
        Prices = new List<dynamic>();
        reader.Read();
        LotName = reader.ReadElementContentAsString("lot_name", "");
        reader.Read();
        while (reader.Name.StartsWith("divisions_"))
        {
            reader.Read();
            var i = reader.ReadElementContentAsString("divisions", "");
            var m = reader.ReadElementContentAsString("match", "");
            var p = reader.ReadElementContentAsString("pay", "");
            Prices.Add(new { ID = i, Match = m, Pay = p });
            reader.Read();
        }

    }

    public void WriteXml(System.Xml.XmlWriter writer)
    {
        throw new NotImplementedException();
    }

    #endregion
}

public class Games  {
    [XmlElement("sports")]
    public Sports Sports { get; set; }
}


[XmlRoot("body")]
public class Body 
{
    [XmlElement("games")]
    public Games Games { get; set; }

}

It uses XmlSerialization because I am not quite sure how to do it with Data Contract Serializer (I know it can be done though). And you might want to add some error handling to the ReadXml method.

PS. You could do it with one class that implements XmlSerializer and emits a lot of dynamic classes :)

.NET XML Deserialization

3 Answers3