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Home Interviews How do you deserialize an XML file into an object? – MindStick
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1 Answers

ICSM Computer
ICSM Computer 07 May 2025
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To deserialize an XML file into an object in C#, you typically use the XmlSerializer class from the System.Xml.Serialization namespace.

Step-by-step: XML to Object

1. Example XML (person.xml):

<Person>
  <Id>1</Id>
  <Name>John Doe</Name>
  <Email>john@example.com</Email>
</Person>

2. Define the C# class:

using System;
using System.Xml.Serialization;

[XmlRoot("Person")]
public class Person
{
    public int Id { get; set; }
    public string Name { get; set; }
    public string Email { get; set; }
}

3. Deserialize the XML:

using System;
using System.IO;
using System.Xml.Serialization;

class Program
{
    static void Main()
    {
        string xmlPath = @"C:\person.xml";

        XmlSerializer serializer = new XmlSerializer(typeof(Person));
        using (FileStream stream = new FileStream(xmlPath, FileMode.Open))
        {
            Person person = (Person)serializer.Deserialize(stream);
            Console.WriteLine($"{person.Id}: {person.Name} - {person.Email}");
        }
    }
}

Notes:

  1. The XML element names must match the class property names (case-sensitive unless overridden with [XmlElement("...")]).
  2. You can deserialize XML strings too using StringReader instead of FileStream.

Deserialize a list of objects from XML into a List<T> in C# using XmlSerializer.

1. Example XML with Multiple <Person> Elements

Save this as people.xml:

<People>
  <Person>
    <Id>1</Id>
    <Name>John Doe</Name>
    <Email>john@example.com</Email>
  </Person>
  <Person>
    <Id>2</Id>
    <Name>Jane Smith</Name>
    <Email>jane@example.com</Email>
  </Person>
</People>

2. Define C# Classes to Match the XML

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

[XmlRoot("People")]
public class People
{
    [XmlElement("Person")]
    public List<Person> Persons { get; set; }
}

public class Person
{
    public int Id { get; set; }
    public string Name { get; set; }
    public string Email { get; set; }
}

3. Deserialize XML into a List of Person

using System;
using System.IO;
using System.Xml.Serialization;

class Program
{
    static void Main()
    {
        string xmlPath = @"C:\people.xml";

        XmlSerializer serializer = new XmlSerializer(typeof(People));
        using (FileStream fs = new FileStream(xmlPath, FileMode.Open))
        {
            People people = (People)serializer.Deserialize(fs);
            foreach (var person in people.Persons)
            {
                Console.WriteLine($"{person.Id}: {person.Name} - {person.Email}");
            }
        }
    }
}

Notes

  1. The root element <People> maps to the People class.
  2. Each <Person> becomes an item in the List<Person>.
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