CODEWITHSUNDEEP
c#xmldeserialization
Özgür Sahin
Özgür Sahin

Reputation: 19

Deserialization of the XML File in C#

I'm trying to deserialize a XML file which Looks like the following:

<xml>
    <classes>
        <class name="EventLog">
            <attribute name="TYPE" type="int"></attribute>
            <attribute name="DESCRIPTION" type="string"></attribute>
            <attribute name="ISSUEDBY" type="string></attribute>
            <attribute name="DATE" type="hr_clock"></attribute>
        </class>
        <class name="test">
            <attribute name="ttt" type="int"></attribute>
            <attribute name="ppp" type="string"></attribute>
            <attribute name="xxx" type="string"></attribute>
            <attribute name="aaa" type="hr_clock"></attribute>
        </class>
    </classes>
    <filters>
        <filter name="COILORDERFILTER">
            <attribute name="COILID" type="string"></attribute> 
            <attribute name="RELIABID" type="string"></attribute>   
        </filter>     
        <filter name="DriveDataFilter">
            <attribute name="DRIVEID" type="string"></attribute>        
        </filter>
    </filters>
</xml>

I need only the classes between the nodes classes. I created following classes for the deserialization:

[Serializable()]
[System.Xml.Serialization.XmlRoot("classes")]
public class ClassCollection
{
    [XmlArray("class")]
    [XmlArrayItem("attribute", typeof(SingleClass))]
    public SingleClass[] singleClass { get; set; }
}

[Serializable()]
public class SingleClass
{
    [System.Xml.Serialization.XmlAttribute("name")]
    public string name { get; set; }

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

class Program
{
    static void Main(string[] args)
    {
        ClassCollection classes = null;
        string path = @"C:\Users\test\Desktop\Eventlog.xml";

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

        StreamReader reader = new StreamReader(path);

        try
        {
            classes = (ClassCollection)serializer.Deserialize(reader);
        }
        catch (InvalidOperationException excep)
        {
            Console.WriteLine(excep.ToString());
        }

        Console.Read();
    }
}

Can anybody say me "what is wrong"?

Upvotes: 1

Views: 831

Answers (3)

Arnaud Develay
Arnaud Develay

Reputation: 3970

If you can't rework your xml, your question is very similar to this one: Ignore outer elements using XmlSerializer

Personnaly, I will use XmlReader instead of XmlSerializer to deserialize your sample XML because it allows to build a model that is not heavily coupled to the XML you are reading.

Upvotes: 0

Marc Gravell
Marc Gravell

Reputation: 1062975

Here's a tip; put your sample xml in a file - my.xml, say, then:

xsd my.xml
xsd my.xsd /classes

This creates my.cs which matches the xml.

Alternatively: you need to understand all the markup; the root is <xml/>, so you need [XmlRoot("xml")], for example. And you almost certainly mean:

    [XmlArray("classes")]
    [XmlArrayItem("class", typeof(SingleClass))]
    public SingleClass[] singleClass { get; set; }

This works, for example:

[XmlRoot("xml")]
public class ClassCollection
{
    [XmlArray("classes")]
    [XmlArrayItem("class", typeof(SingleClass))]
    public SingleClass[] singleClass { get; set; }
}
public class SingleClass
{
    [XmlElement("attribute")]
    public List<Attribute> Attributes { get; set; }

    [XmlAttribute("name")]
    public string Name { get; set; }
}
public class Attribute {
    [XmlAttribute("name")]
    public string name { get; set; }

    [XmlAttribute("type")]
    public string type { get; set; }
}

Upvotes: 4

Sauleil
Sauleil

Reputation: 2603

Not sure about your problem since there are no errors

but 1st:

[XmlRoot("classes")]

It means that the first XML node should be "classes" and it's "xml" right now from your example. So create a new object:

[XmlRoot("xml")]
class MyRoot {
    //...

Also, usually you'll want to have something like this as your first line for the default serializer to work I think:

<?xml version="1.0" encoding="utf-8" ?>

Also, I think you have some errors in your dataclass definition. You should comment some stuff and do some tests with less data in order to understand properly what's happening.

Upvotes: 0

Related Questions