How to Deserialize XML document

Question

Hi, How do I Deserialize this XML document:

<?xml version="1.0" encoding="utf-8"?>
<Cars>
  <Car>
    <StockNumber>1020</StockNumber>
    <Make>Nissan</Make>
    <Model>Sentra</Model>
  </Car>
  <Car>
    <StockNumber>1010</StockNumber>
    <Make>Toyota</Make>
    <Model>Corolla</Model>
  </Car>
  <Car>
    <StockNumber>1111</StockNumber>
    <Make>Honda</Make>
    <Model>Accord</Model>
  </Car>
</Cars>

I have this:

[Serializable()]
public class Car
{
    [System.Xml.Serialization.XmlElementAttribute("StockNumber")]
    public string StockNumber{ get; set; }

    [System.Xml.Serialization.XmlElementAttribute("Make")]
    public string Make{ get; set; }

    [System.Xml.Serialization.XmlElementAttribute("Model")]
    public string Model{ get; set; }
}

.

[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
    [XmlArrayItem(typeof(Car))]
    public Car[] Car { get; set; }

}

.

public class CarSerializer
{
    public Cars Deserialize()
    {
        Cars[] cars = null;
        string path = HttpContext.Current.ApplicationInstance.Server.MapPath("~/App_Data/") + "cars.xml";

        XmlSerializer serializer = new XmlSerializer(typeof(Cars[]));

        StreamReader reader = new StreamReader(path);
        reader.ReadToEnd();
        cars = (Cars[])serializer.Deserialize(reader);
        reader.Close();

        return cars;
    }
}

that don't seem to work :-(

Answer 1

See if this helps:

[Serializable()]
[System.Xml.Serialization.XmlRootAttribute("Cars", Namespace = "", IsNullable = false)]
public class Cars
{
    [XmlArrayItem(typeof(Car))]
    public Car[] Car { get; set; }
}

.

[Serializable()]
public class Car
{
    [System.Xml.Serialization.XmlElement()]
    public string StockNumber{ get; set; }

    [System.Xml.Serialization.XmlElement()]
    public string Make{ get; set; }

    [System.Xml.Serialization.XmlElement()]
    public string Model{ get; set; }
}

And failing that use the xsd.exe program that comes with visual studio to create a schema document based on that xml file, and then use it again to create a class based on the schema document.

Answer 2

Here's a working version. I changed the XmlElementAttribute labels to XmlElement because in the xml the StockNumber, Make and Model values are elements, not attributes. Also I removed the reader.ReadToEnd(); (that function reads the whole stream and returns a string, so the Deserialze() function couldn't use the reader anymore...the position was at the end of the stream). I also took a few liberties with the naming :).

Here are the classes:

[Serializable()]
public class Car
{
    [System.Xml.Serialization.XmlElement("StockNumber")]
    public string StockNumber { get; set; }

    [System.Xml.Serialization.XmlElement("Make")]
    public string Make { get; set; }

    [System.Xml.Serialization.XmlElement("Model")]
    public string Model { get; set; }
}


[Serializable()]
[System.Xml.Serialization.XmlRoot("CarCollection")]
public class CarCollection
{
    [XmlArray("Cars")]
    [XmlArrayItem("Car", typeof(Car))]
    public Car[] Car { get; set; }
}

The Deserialize function:

CarCollection cars = null;
string path = "cars.xml";

XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));

StreamReader reader = new StreamReader(path);
cars = (CarCollection)serializer.Deserialize(reader);
reader.Close();

And the slightly tweaked xml (I needed to add a new element to wrap <Cars>...Net is picky about deserializing arrays):

<?xml version="1.0" encoding="utf-8"?>
<CarCollection>
<Cars>
  <Car>
    <StockNumber>1020</StockNumber>
    <Make>Nissan</Make>
    <Model>Sentra</Model>
  </Car>
  <Car>
    <StockNumber>1010</StockNumber>
    <Make>Toyota</Make>
    <Model>Corolla</Model>
  </Car>
  <Car>
    <StockNumber>1111</StockNumber>
    <Make>Honda</Make>
    <Model>Accord</Model>
  </Car>
</Cars>
</CarCollection>

Answer 3

How about you just save the xml to a file, and use xsd?

1. Write the file to disk (I named it foo.xml)
2. Generate the xsd: xsd foo.xml
3. Generate the C#: xsd foo.xsd /classes

Et voila - and C# code file that should be able to read the data via XmlSerializer:

    XmlSerializer ser = new XmlSerializer(typeof(Cars));
    Cars cars;
    using (XmlReader reader = XmlReader.Create(path))
    {
        cars = (Cars) ser.Deserialize(reader);
    }

(include the generated foo.cs in the project)

Answer 4

The following snippet should do the trick (and you can ignore most of the serialization attributes):

public class Car
{
  public string StockNumber { get; set; }
  public string Make { get; set; }
  public string Model { get; set; }
}

[XmlRootAttribute("Cars")]
public class CarCollection
{
  [XmlElement("Car")]
  public Car[] Cars { get; set; }
}

...

using (TextReader reader = new StreamReader(path))
{
  XmlSerializer serializer = new XmlSerializer(typeof(CarCollection));
  return (CarCollection) serializer.Deserialize(reader);
}