XML Embedded Resource to Object

You can create an XML file which has configurations, constants or simple data in it, and in its file properties set Build Action to Embedded Resource.
Then a related model with XmlAttributes is required.

Notice the differences among XmlRoot, XmlAttribue, XmlElement, or this will cause parse errors.

TeamModel.cs

    [XmlRoot("TeamModel")]
    public class TeamModel
    {
        private string _name = string.Empty;

        [XmlAttribute("Name")]
        public string Name
        {
            get { return _name; }
            set { _name = value; }
        }

        private List<MemberModel> aList = new List<MemberModel>();
        [XmlElement("Member")]
        public List<MemberModel> AList
        {
            get { return aList; }
            set { aList = value; }
        }
    }

    public class MemberModel
    {
        private string _firstName = string.Empty;
        [XmlAttribute("FirstName")]
        public string FirstName
        {
            get { return _firstName; }
            set { _firstName = value; }
        }

        private string _lastName = string.Empty;
        [XmlAttribute("LastName")]
        public string LastName
        {
            get { return _lastName; }
            set { _lastName = value; }
        }
    }

TeamModelConfig.xml

<?xml version="1.0" encoding="utf-8"?>
<ArrayOfTeamModel >
  <TeamModel Name="MyTeam">
    <Member FirstName="Zhang" LastName="San"/>
    <Member FirstName="Li" LastName="Si"></Member>
  </TeamModel>
</ArrayOfTeamModel>

Method for XML parse

public T ParseEmbeddedXml<T>(string resourceName)
{
    var result = default(T);
    //use reflection to get executing assembly and its name
    using (Stream stream = Assembly.GetExecutingAssembly().GetManifestResourceStream(
      Assembly.GetExecutingAssembly().GetName().Name.ToString()+"."+resourceName))
    {
        using (StreamReader reader = new StreamReader(stream))
        {
            XmlSerializer xmlSerializer = XmlSerializer.FromTypes(new[] { typeof(T) }).FirstOrDefault();//new XmlSerializer(typeof(T));
            var tempResult = xmlSerializer.Deserialize(reader);
            if (tempResult is T)
                result = (T)tempResult;
        }
    }
    return result;
}