Deserialization of the XML File in C#

Question

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"?

At first impact, I'd say that you are trying to deserialize the classes nodes, but you give to the serializer the full xml — Matteo Umili, May 26 '15 at 12:17
Note that you don't need the `Serializable` attributes for this kind of serialization... http://stackoverflow.com/questions/392431/why-doesnt-the-xmlserializer-need-the-type-to-be-marked-serializable — Baldrick, May 26 '15 at 12:22

Marc Gravell · Answer 1 · 2015-05-26T12:26:20.623

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; }
}

score 0 · Answer 2 · answered May 26 '15 at 12:22

0

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.

answered May 26 '15 at 12:22

Sauleil

2,573
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The "usually you'll want to have" is unnecessary - the xml description header is not compulsory – Marc Gravell May 26 '15 at 12:27
I remember having some issues with the XmlSerializer when I didn't have that header. I'm just writing that statement from memory but you're probably right and it's not mandatory. But I think it's a good practice anyway. – Sauleil May 27 '15 at 12:41

score 0 · Answer 3 · edited May 23 '17 at 11:51

0

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.

edited May 23 '17 at 11:51

Community

1
1

answered May 26 '15 at 12:35

Arnaud Develay

3,920
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Deserialization of the XML File in C#

3 Answers3