Deserializing Xml to List<T> - xmlns='' was not expected

Question

I have problems with deserializing XML to the List of locations objects.
Given the following XML:

<?xml version="1.0" encoding="utf-8"?>
    <locations>
        <location id="1">
            <level name="3" complete="True" stars="1" firstMisson="True" secondMission="False" thridMission="False" />
        </location>
        <location id="2">
           <level name="4" complete="True" stars="3" firstMisson="True" secondMission="True" thridMission="True" />
        </location>
    </locations>

And the following class:

    [System.Serializable]
    [XmlRoot(ElementName = "level")]
    public class Level
    {
        [XmlAttribute(AttributeName = "name")]
        public string Name { get; set; }
        [XmlAttribute(AttributeName = "complete")]
        public string Complete { get; set; }
        [XmlAttribute(AttributeName = "stars")]
        public string Stars { get; set; }
        [XmlAttribute(AttributeName = "firstMisson")]
        public string FirstMisson { get; set; }
        [XmlAttribute(AttributeName = "secondMission")]
        public string SecondMission { get; set; }
        [XmlAttribute(AttributeName = "thridMission")]
        public string ThridMission { get; set; }
    }
    [System.Serializable]
    [XmlRoot(ElementName = "location")]
    public class Location
    {
        [XmlElement(ElementName = "level")]
        public Level Level { get; set; }
        [XmlAttribute(AttributeName = "id")]
        public string Id { get; set; }
    }
    [System.Serializable]
    [XmlRoot(ElementName = "locations")]
    public class Locations
    {
        [XmlElement(ElementName = "location")]
        public Location Location { get; set; }

        public List<Locations> LocDb = new List<Locations>();
    }
    [System.Serializable]
    [XmlRoot(ElementName = "xml")]
    public class Xml
    {
        [XmlElement(ElementName = "locations")]
        public Locations Locations { get; set; }
    }

Deserialization Method

     public List<Locations> locDB = new List<Locations>();

     public static void LoadData()
        {
            string filepath = Application.dataPath + @"/XML/GameXMLdata.xml";

            var xmlSerializer = new XmlSerializer(locDB.GetType());
            var stream = File.Open(filepath, FileMode.Open);


            locDB = (List<Locations>)xmlSerializer.Deserialize(stream);//locations xmlns=''> was not expected

            stream.Close();

            Debug.Log(locDB[1].Location.Id);

         }

So how do I deserialize XML to List of location objects?

Many thanks for help.

Answer 1

You need only two classes:

[XmlType("level")]
public class Level
{
    [XmlAttribute("name")]
    public string Name { get; set; }
    [XmlAttribute("complete")]
    public string Complete { get; set; }
    [XmlAttribute("stars")]
    public string Stars { get; set; }
    [XmlAttribute("firstMisson")]
    public string FirstMisson { get; set; }
    [XmlAttribute("secondMission")]
    public string SecondMission { get; set; }
    [XmlAttribute("thridMission")]
    public string ThridMission { get; set; }
}

[XmlType("location")]
public class Location
{
    [XmlElement("level")]
    public Level Level { get; set; }
    [XmlAttribute("id")]
    public string Id { get; set; }
}

The important is to apply XmlType attribute to the class Location rather than XmlRoot .

Now you can deserialize your XML to List<Location> like this:

var xmlSerializer = new XmlSerializer(typeof(List<Location>), 
    new XmlRootAttribute("locations"));
List<Location> locations;
using (var stream = File.OpenRead(filepath))
    locations = (List<Location>)xmlSerializer.Deserialize(stream);

The trick is to specify the root element name ("locations" in your case) using the XmlSerializer(Type, XmlRootAttribute) constructor overload.

Answer 2

I had a similar problem on a class property and solved it using a different approach. Since in my scenario I wasn't able to touch the XmlSerializer construction logic, this is what I did (using the types from OP for reference):

[XmlRoot("locations")]
public sealed class LocationCollection : Collection<Location>
{
}

With the custom collection class, I'm able to set the root at design time and still use a standard XmlSeriaizer instance.