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Home/ Questions/Q 5932897
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T14:53:54+00:00 2026-05-22T14:53:54+00:00

I am trying to deserialize an XML file to a class. The XML file

  • 0

I am trying to deserialize an XML file to a class. The XML file looks like this:

<?xml version="1.0" encoding="utf-8"?>
<locations>
<location id="0">
    <name>park</name>
    <temperature>5</temperature>
    <wind>26</wind>
    <weather_text_SI>sunny</weather_text_SI>
    <visibility></visibility>
    <latitude>46.4527</latitude>
    <longitude>15.334</longitude>
    <elevation>1517</elevation>
</location>
</locations>

The class that I want to deserialize it to is:

[XmlRootAttribute("locations")]
public class SnowPark
{
    public SnowPark()
    {
    }

    private int id;
    [XmlAttribute("id")] 
    public int Id
    {
        get { return id; }
        set { id = value; }
    }

    private string name;
    [XmlElement("name")]
    public string Name
    {
        get { return name; }
        set { name = value; }
    }

    private int temperature;
    [XmlElement("temperature")]
    public int Temperature
    {
        get { return temperature; }
        set { temperature = value; }
    }

    private int wind;
    [XmlElement("wind")]
    public int Wind
    {
        get { return wind; }
        set { wind = value; }
    }

    private string weatherText;
    [XmlElement("weather_text_SI")]
    public string WeatherText
    {
        get { return weatherText; }
        set { weatherText = value; }
    }

    private double latitude;
    [XmlElement("latitude")]
    public double Latitude
    {
        get { return latitude; }
        set { latitude = value; }
    }

    private double longitude;
    [XmlElement("longitude")]
    public double Longitude
    {
        get { return longitude; }
        set { longitude = value; }
    }

    private int elevation;
    [XmlElement("elevation")]
    public int Elevation
    {
        get { return elevation; }
        set { elevation = value; }
    }
}

I try to deserialize the XML file

XmlSerializer deserializer = new XmlSerializer(typeof(List<SnowPark>));
TextReader textReader = new StreamReader(@"file.xml");
List<SnowPark> parks;
parkss = (List<SnowPark>)deserializer.Deserialize(textReader);
textReader.Close();

However I get an exception:

There is an error in XML document (2, 2).

and an inner exception:

<locations xmlns=''> was not expected.

No luck finding the solution so far. Help appreciated.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-22T14:53:55+00:00Added an answer on May 22, 2026 at 2:53 pm

    The XmlRootAttribute doesn’t apply since you are serialising a list of then, not an individual item; this also means your XML is one layer further-out than needed.

    IMO, your easiest option here is:

    [XmlRoot("locations")]
    public class Locations
    {
        [XmlElement("location")]
        public List<SnowPark> Parks {get;set;}
    }
    

    and deserialize a Locations object, using typeof(Locations) to initialisers the XmlSerializer

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