Я написал следующий пример кода, чтобы сохранить слегка сложный объект FamilyTreeFile
в XML и восстановить его в исходной форме.
public class XmlSerializationTest
{
const string FileName = @"FamilyTree.xml";
public void Run()
{
var rootMember = new Member() { Name = "Johny", Parent = null };
var member1 = new Member() { Name = "Andy", Parent = rootMember };
var member2 = new Member() { Name = "Adam", Parent = rootMember };
var member3 = new Member() { Name = "Andrew", Parent = rootMember };
var member4 = new Member() { Name = "Davis", Parent = member2 };
var member5 = new Member() { Name = "Simon", Parent = member4 };
rootMember.FamilyTree = new GenericCollection();
rootMember.FamilyTree.Add(member1);
rootMember.FamilyTree.Add(member2);
rootMember.FamilyTree.Add(member3);
member2.FamilyTree = new GenericCollection();
member2.FamilyTree.Add(member4);
member4.FamilyTree = new GenericCollection();
member4.FamilyTree.Add(member5);
var familyTree = new GenericCollection() { rootMember };
IFamilyTreeFile file = new FamilyTreeFile()
{
FamilyTree = familyTree
};
Serialize(file);
file = Deserialize();
}
public void Serialize(IFamilyTreeFile obj)
{
var xmlSerializer = new XmlSerializer(typeof(FamilyTreeFile));
using (TextWriter writer = new StreamWriter(FileName))
{
xmlSerializer.Serialize(writer, obj);
}
}
public IFamilyTreeFile Deserialize()
{
XmlSerializer serializer = new XmlSerializer(typeof(FamilyTreeFile));
using (Stream stream = File.Open(FileName, FileMode.Open))
{
return (IFamilyTreeFile)serializer.Deserialize(stream);
}
}
}
public interface IMember
{
string Name { get; set; }
IMember Parent { get; set; }
GenericCollection FamilyTree { get; set; }
}
[Serializable]
public class Member : IMember
{
[XmlAttribute]
public string Name { get; set; }
[XmlIgnore]
public IMember Parent { get; set; }
public GenericCollection FamilyTree { get; set; }
public Member()
{
//FamilyTree = new GenericCollection();
}
}
[Serializable]
public class GenericCollection : List<IMember>, IXmlSerializable
{
public System.Xml.Schema.XmlSchema GetSchema()
{
return null;
}
public void ReadXml(XmlReader reader)
{
reader.MoveToContent();
if (reader.Name == "FamilyTree")
{
do
{
reader.Read();
if (reader.Name == "Member" && reader.IsStartElement())
{
Type type = System.Reflection.Assembly.GetExecutingAssembly().GetTypes()
.Where(x => x.Name == reader.Name)
.FirstOrDefault();
if (type != null)
{
var xmlSerializer = new XmlSerializer(type);
var member = (IMember)xmlSerializer.Deserialize(reader);
this.Add(member);
}
}
if (reader.Name == "FamilyTree" && reader.NodeType == System.Xml.XmlNodeType.EndElement)
break;
}
while (!reader.EOF);
}
}
public void WriteXml(XmlWriter writer)
{
foreach (IMember rule in this)
{
var namespaces = new XmlSerializerNamespaces();
namespaces.Add(String.Empty, String.Empty);
XmlSerializer xmlSerializer = new XmlSerializer(rule.GetType());
xmlSerializer.Serialize(writer, rule, namespaces);
}
}
}
public interface IFamilyTreeFile
{
GenericCollection FamilyTree { get; set; }
}
public class FamilyTreeFile : IFamilyTreeFile
{
public GenericCollection FamilyTree { get; set; }
}
Образец кода генерирует следующий XML-файл, который точно соответствует моим потребностям, но я не могу прочитать его с помощью метода ReadXml.
<?xml version="1.0" encoding="utf-8"?>
<FamilyTreeFile xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:xsd="http://www.w3.org/2001/XMLSchema">
<FamilyTree>
<Member Name="Johny">
<FamilyTree>
<Member Name="Andy" />
<Member Name="Adam">
<FamilyTree>
<Member Name="Davis">
<FamilyTree>
<Member Name="Simon" />
</FamilyTree>
</Member>
</FamilyTree>
</Member>
<Member Name="Andrew" />
</FamilyTree>
</Member>
</FamilyTree>
</FamilyTreeFile>
Мне нужна помощь в том, как я могу эффективно восстановить его?
ДОБАВЛЕНО
При добавлении новой коллекции Notes
в IMember
public interface IMember
{
string Name { get; set; }
IMember Parent { get; set; }
GenericCollection FamilyTree { get; set; }
List<Note> Notes { get; set; }
}
[Serializable]
public class Note
{
[XmlAttribute]
public string Text { get; set; }
}
Реализация этого свойства в Member
классе
[XmlArray("Notes")]
public List<Note> Notes { get; set; }
Я не могу десериализовать информацию Notes в этой строке.
var member = (IMember)xmlSerializer.Deserialize(reader);
Нет ли простого способа десериализации с помощью XmlSerializer или какой-либо инфраструктуры, которая сама все обрабатывает?