XML, Xsd, C#

Question

hello every one,

how can i deserelize an xsd generated class in order to use it?

here is my class. I have to show the elements in a combo box.

thanks for support :) 

//------------------------------------------------------------------------------
// <auto-generated>
//     Ce code a été généré par un outil.
//     Version du runtime :4.0.30319.42000
//
//     Les modifications apportées à ce fichier peuvent provoquer un comportement incorrect et seront perdues si
//     le code est régénéré.
// </auto-generated>
//------------------------------------------------------------------------------

using System.Xml.Serialization;

// 
// This source code was auto-generated by xsd, Version=4.7.2046.0.
// 

namespace Xml_read.Model
{
    /// <remarks/>
    [System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.7.2046.0")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)]
[System.Xml.Serialization.XmlRootAttribute(Namespace="", IsNullable=false)]
public partial class voitures {
    
    private voituresVoiture[] itemsField;
    
    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("voiture", Form=System.Xml.Schema.XmlSchemaForm.Unqualified)]
    public voituresVoiture[] Items {
        get {
            return this.itemsField;
        }
        set {
            this.itemsField = value;
        }
    }
}

/// <remarks/>
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.7.2046.0")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)]
public partial class voituresVoiture {
    
    private string nb_voituresField;
    
    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute(Form=System.Xml.Schema.XmlSchemaForm.Unqualified)]
    public string nb_voitures {
        get {
            return this.nb_voituresField;
        }
        set {
            this.nb_voituresField = value;
        }
    }
}
}

Answer 1

It depends a little bit on the form of XML, whether it is a file or a string or a stream..

// From file.
string fileName = @"C:\Path\YourFile.Xml";
XmlSerializer serializer = new XmlSerializer(typeof(voitures));
using (FileStream fileStream = new FileStream(fileName, FileMode.Open))
{
    voitures result = (voitures)serializer.Deserialize(fileStream);
}

// From string.
string sampleString = @"<voitures>..</voitures>"; // Correct XML needed ;)
XmlSerializer serializer = new XmlSerializer(typeof(voitures));
using (TextReader reader = new StringReader(sampleString))
{
    voitures result = (voitures )serializer.Deserialize(reader);
}

See Xml​Serializer.​Deserialize.

Answer 2

hello sir, 

it is a file.

it actualy works with the first sollution. thank you! :)