Is it possible to deserialize XML into List?

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

Yes, it will serialize and deserialize a List<>. Just make sure you use the [XmlArray] attribute if in doubt.

[Serializable]
public class A
{
    [XmlArray]
    public List<string> strings;
}

This works with both Serialize() and Deserialize().

Answer 2

Not sure about List<T> but Arrays are certainly do-able. And a little bit of magic makes it really easy to get to a List again.

public class UserHolder {
   [XmlElement("list")]
   public User[] Users { get; set; }

   [XmlIgnore]
   public List<User> UserList { get { return new List<User>(Users); } }
}

Answer 3

How about

XmlSerializer xs = new XmlSerializer(typeof(user[]));
using (Stream ins = File.Open(@"c:\some.xml", FileMode.Open))
foreach (user o in (user[])xs.Deserialize(ins))
   userList.Add(o);    

Not particularly fancy but it should work.

Answer 4

I think I have found a better way. You don't have to put attributes into your classes. I've made two methods for serialization and deserialization which take generic list as parameter.

Take a look (it works for me):

private void SerializeParams<T>(XDocument doc, List<T> paramList)
    {
        System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(paramList.GetType());

        System.Xml.XmlWriter writer = doc.CreateWriter();

        serializer.Serialize(writer, paramList);

        writer.Close();           
    }

private List<T> DeserializeParams<T>(XDocument doc)
    {
        System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(typeof(List<T>));

        System.Xml.XmlReader reader = doc.CreateReader();

        List<T> result = (List<T>)serializer.Deserialize(reader);
        reader.Close();

        return result;
    }

So you can serialize whatever list you want! You don't need to specify the list type every time.

        List<AssemblyBO> list = new List<AssemblyBO>();
        list.Add(new AssemblyBO());
        list.Add(new AssemblyBO() { DisplayName = "Try", Identifier = "243242" });
        XDocument doc = new XDocument();
        SerializeParams<T>(doc, list);
        List<AssemblyBO> newList = DeserializeParams<AssemblyBO>(doc);

Answer 5

Yes, it does deserialize to List<>. No need to keep it in an array and wrap/encapsulate it in a list.

public class UserHolder
{
    private List<User> users = null;

    public UserHolder()
    {
    }

    [XmlElement("user")]
    public List<User> Users
    {
        get { return users; }
        set { users = value; }
    }
}

Deserializing code,

XmlSerializer xs = new XmlSerializer(typeof(UserHolder));
UserHolder uh = (UserHolder)xs.Deserialize(new StringReader(str));

Answer 6

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.