Is it possible to deserialize XML into List<T>?

Question

Given the following XML:

<?xml version="1.0"?> <user_list>    <user>       <id>1</id>       <name>Joe</name>    </user>    <user>       <id>2</id>       <name>John</name>    </user> </user_list> 

And the following class:

public class User {    [XmlElement("id")]    public Int32 Id { get; set; }     [XmlElement("name")]    public String Name { get; set; } } 

Is it possible to use XmlSerializer to deserialize the xml into a List<User> ? If so, what type of additional attributes will I need to use, or what additional parameters do I need to use to construct the XmlSerializer instance?

An array ( User[] ) would be acceptable, if a bit less preferable.

Answer 1

You can encapsulate the list trivially:

using System; using System.Collections.Generic; using System.Xml.Serialization;  [XmlRoot("user_list")] public class UserList {     public UserList() {Items = new List<User>();}     [XmlElement("user")]     public List<User> Items {get;set;} } public class User {     [XmlElement("id")]     public Int32 Id { get; set; }      [XmlElement("name")]     public String Name { get; set; } }  static class Program {     static void Main()     {         XmlSerializer ser= new XmlSerializer(typeof(UserList));         UserList list = new UserList();         list.Items.Add(new User { Id = 1, Name = "abc"});         list.Items.Add(new User { Id = 2, Name = "def"});         list.Items.Add(new User { Id = 3, Name = "ghi"});         ser.Serialize(Console.Out, list);     } } 

Answer 2

If you decorate the User class with the XmlType to match the required capitalization:

[XmlType("user")] public class User {    ... } 

Then the XmlRootAttribute on the XmlSerializer ctor can provide the desired root and allow direct reading into List<>:

    // e.g. my test to create a file     using (var writer = new FileStream("users.xml", FileMode.Create))     {         XmlSerializer ser = new XmlSerializer(typeof(List<User>),               new XmlRootAttribute("user_list"));         List<User> list = new List<User>();         list.Add(new User { Id = 1, Name = "Joe" });         list.Add(new User { Id = 2, Name = "John" });         list.Add(new User { Id = 3, Name = "June" });         ser.Serialize(writer, list);     } 

...

    // read file     List<User> users;     using (var reader = new StreamReader("users.xml"))     {         XmlSerializer deserializer = new XmlSerializer(typeof(List<User>),               new XmlRootAttribute("user_list"));         users = (List<User>)deserializer.Deserialize(reader);     } 

Credit: based on answer from YK1.