Answered by:
XML, Xsd, C#

Question
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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; } } } }
Answers
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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 XmlSerializer.Deserialize.- Marked as answer by Bil59650 Thursday, April 11, 2019 12:27 PM
All replies
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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 XmlSerializer.Deserialize.- Marked as answer by Bil59650 Thursday, April 11, 2019 12:27 PM
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