build classes from xml file

Question

First of all, i'm sorry, i know my question will be very basic and stuff but please i would really appreciate some help, i'm under lots of pressure, so making smart comments and being ironic wouldn't help

I have this code: At least i'm trying. I have this xml file, i need to save all it's attributes in a class so i can just duplicate it whenever i need to.

I don't know how to.

public static void firstFileXml(string sXml1)
{
    var root = XElement.Load(@"sXml1");
    var controlElementsFirst = root.Descendants("books");
}

The XML file has attributes like: label, text, label_w etc. I need a function or somethign that will allow me to enter like: explore(xmlLocation) and to do the rest. Because i need to do it for a few xml files

I need to build a class that will allow me to read it. Suppose i have this file of xml:

WAS

    <books>
        <book label='' page='' intro =''/>
        <book label='' page='' intro =''/>
        <book label='' page='' intro =''/>
    </books>

IS

<SHEET>
<books>
<book label='1' page='1' intro='1'/>
<book label='2' page='2' intro='2'/>
<book label='3' page='3' intro='3'/>
</books>
</SHEET>

And so on. I need first to read this xml file then store the book attributes in a class, so i can use it for hundreds of books later

Code:

using System;
using System.Collections.Generic;
using System.IO;
using System.Xml.Serialization;

static class Program
{
    static void Main()
    {
        // get the xml as a string; note that we could also use
        // Deserialize(Stream) to process a FileStream, but this
        // works fine
        string xml = File.ReadAllText(@"C:\Users\books.xml");
        var ser = new XmlSerializer(typeof(BookRoot));
        var root = (BookRoot)ser.Deserialize(new StringReader(xml));
        foreach (var book in root.Books)
        {
            Console.WriteLine("label: " + book.Label);
            Console.WriteLine("page: " + book.Page);
            Console.WriteLine("intro: " + book.Intro);
            Console.ReadLine();
        }
    }
}
[XmlRoot("SHEET")]
public class BookRoot
{
    private readonly List<Book> books = new List<Book>();
    [XmlArray("books"), XmlArrayItem("book")]
    public List<Book> Books { get { return books; } }
}
public class Book
{
    [XmlAttribute("label")]
    public string Label { get; set; }
    [XmlAttribute("page")]
    public string Page { get; set; }
    [XmlAttribute("intro")]
    public string Intro { get; set; }
}

Not really because they have nodes while i'm using mostly attributes. Please i would really appreciate some sample algorithm or tiny code, just to have an idea — Ani Ani
– Ani Ani, Commented Sep 2, 2014 at 7:58
i fixed the xml file structure, now it's ok, anyways the xml itself is ok — Ani Ani
– Ani Ani, Commented Sep 2, 2014 at 8:01

Marc Gravell · Accepted Answer · 2014-09-02 09:06:59Z

4

At the command line:

xsd yourfile.xml
xsd /c yourfile.xsd

This will create C# classes using the xml as a template (via schema inference); you can then use those classes with XmlSerializer, i.e.

var ser = new XmlSerializer(typeof(YourRootType));
YourRootType root = (YourRootType)ser.Deserialize(source);
Console.WriteLine(root.Foo);
foreach(Bar bar in root.Bars) {
    Console.WriteLine(bar.Name);
    Console.WriteLine(bar.Id);
}

For example, with:

<books>
<book label='' page='' intro=''/>
<book label='' page='' intro=''/>
<book label='' page='' intro=''/>
</books>

this generates the schema:

<?xml version="1.0" encoding="utf-8"?>
<xs:schema id="books" xmlns="" xmlns:xs="http://www.w3.org/2001/XMLSchema"
    xmlns:msdata="urn:schemas-microsoft-com:xml-msdata">
  <xs:element name="books" msdata:IsDataSet="true" msdata:Locale="en-US">
    <xs:complexType>
      <xs:choice minOccurs="0" maxOccurs="unbounded">
        <xs:element name="book">
          <xs:complexType>
            <xs:attribute name="label" type="xs:string" />
            <xs:attribute name="page" type="xs:string" />
            <xs:attribute name="intro" type="xs:string" />
          </xs:complexType>
        </xs:element>
      </xs:choice>
    </xs:complexType>
  </xs:element>
</xs:schema>

and then the C# code:

//------------------------------------------------------------------------------
// <auto-generated>
//     This code was generated by a tool.
//     Runtime Version:4.0.30319.34014
//
//     Changes to this file may cause incorrect behavior and will be lost if
//     the code is regenerated.
// </auto-generated>
//------------------------------------------------------------------------------

using System.Xml.Serialization;

// 
// This source code was auto-generated by xsd, Version=4.0.30319.18020.
// 


/// <remarks/>
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.18020")]
[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 books {

    private booksBook[] itemsField;

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("book", Form=System.Xml.Schema.XmlSchemaForm.Unqualified)]
    public booksBook[] Items {
        get {
            return this.itemsField;
        }
        set {
            this.itemsField = value;
        }
    }
}

/// <remarks/>
[System.CodeDom.Compiler.GeneratedCodeAttribute("xsd", "4.0.30319.18020")]
[System.SerializableAttribute()]
[System.Diagnostics.DebuggerStepThroughAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType=true)]
public partial class booksBook {

    private string labelField;

    private string pageField;

    private string introField;

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string label {
        get {
            return this.labelField;
        }
        set {
            this.labelField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string page {
        get {
            return this.pageField;
        }
        set {
            this.pageField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute()]
    public string intro {
        get {
            return this.introField;
        }
        set {
            this.introField = value;
        }
    }
}

which works fine with XmlSerializer, and will correctly deserialize your data, for example:

var ser = new XmlSerializer(typeof(books));
var root = (books)ser.Deserialize(new StringReader(xml));
foreach(var book in root.Items)
{
    Console.WriteLine("label: " + book.label);
    Console.WriteLine("page: " + book.page);
    Console.WriteLine("intro: " + book.intro);
}

Note that you could do it by hand a lot easier:

[XmlRoot("books")]
public class BookRoot {
    private readonly List<Book> books = new List<Book>();
    [XmlElement("book")]
    public List<Book> Books { get { return books; } }
}
public class Book {
    [XmlAttribute("label")]
    public string Label {get;set;}
    [XmlAttribute("page")]
    public string Page {get;set;}
    [XmlAttribute("intro")]
    public string Intro {get;set;}
}

for example:

var ser = new XmlSerializer(typeof(BookRoot));
var root = (BookRoot)ser.Deserialize(new StringReader(xml));
foreach(var book in root.Books)
{
    Console.WriteLine("label: " + book.Label);
    Console.WriteLine("page: " + book.Page);
    Console.WriteLine("intro: " + book.Intro);
}

With the most recent edit that adds a level into the XML, the attributes need tweaking a bit - to change the root element name, and to add an additional level (books/book => SHEET/books/book):

[XmlRoot("SHEET")]
public class BookRoot
{
    private readonly List<Book> books = new List<Book>();
    [XmlArray("books"), XmlArrayItem("book")]
    public List<Book> Books { get { return books; } }
}

edited Sep 2, 2014 at 9:06

answered Sep 2, 2014 at 7:58

Marc Gravell

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33 Comments

Ani Ani Over a year ago

Thank you for your help, could you take a look at my structure of xml? besides, i'm working on visual studio

Marc Gravell Over a year ago

@AniAni your xml isn't actually xml, but I'll edit in what it does with something based on that. The xsd.exe tool is a freely available part of the .NET sdk

Ani Ani Over a year ago

Thanks, but it would be better for me if it'S not a tool, well i wish i could understand it but it's difficult for me

Marc Gravell Over a year ago

@AniAni you say "I need to build a class that will allow me to read it."; that is exactly and completely what the xsd.exe tool does. I'm not exactly sure what the issue is here... but you can do it by hand too: see the much shorter (but equivalent) hand-coded version at the end.

AFract Over a year ago

xsd.exe is the way to go, at least for first step (generation of XSD from XML). You could also try xsd2code.codeplex.com which offers a few more options than xsd.exe for class generation from .xsd file.

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