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Deserialize XML to List using additional class wrapper
You can use XmlSerializer
to deserialize XML to List<T>
by using an additional class to encapsulate the list.
Packaging class
Create a class that encapsulates a list, for example:
<code class="language-csharp">[XmlRoot("user_list")] public class UserList { public UserList() { Items = new List<User>(); } [XmlElement("user")] public List<User> Items { get; set; } }</code>
User class
Define the User
class as before:
<code class="language-csharp">public class User { [XmlElement("id")] public Int32 Id { get; set; } [XmlElement("name")] public String Name { get; set; } }</code>
Deserialization code
Deserialize the XML using the following code:
<code class="language-csharp">using System.Xml.Serialization; XmlSerializer ser = new XmlSerializer(typeof(UserList)); UserList list = (UserList)ser.Deserialize(new XmlTextReader("users.xml"));</code>
This will deserialize the XML into a UserList
class, which contains a list of User
objects.
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