XmlIncludeAttribute.Type プロパティ

.NET Framework クラス ライブラリ リファレンス

XmlIncludeAttribute.Type プロパティ

含めるオブジェクトの型を取得または設定します。

名前空間: System.Xml.Serialization
アセンブリ: System.Xml (system.xml.dll 内)
構文

Visual Basic (宣言)

Public Property Type As
 Type

Visual Basic (使用法)

Dim instance As XmlIncludeAttribute
Dim value As Type

value = instance.Type

instance.Type = value

C#

public Type Type { get; set;
 }

C++

public:
property Type^ Type {
    Type^ get ();
    void set (Type^ value);
}

J#

/** @property */
public Type get_Type ()

/** @property */
public void set_Type (Type value)

JScript

public function get Type
 () : Type

public function set Type
 (value : Type)

プロパティ値
含めるオブジェクトの Type。

使用例

Employee オブジェクトの配列を返す Employees フィールドを含む Group クラスを定義する例を次に示します。この例では、Employee クラスから Manager クラスを派生させ、XmlIncludeAttribute を Employee クラスに適用します。Group オブジェクトが作成されると、Manager オブジェクトを Employee 配列に挿入します。最後に、Group オブジェクトをシリアル化します。

Visual Basic

Imports System
Imports System.IO
Imports System.Xml.Serialization
Imports Microsoft.VisualBasic


Public Class Group
    Public Employees() As Employee
End Class

' Instruct the XmlSerializer to accept Manager types as well.
<XmlInclude(GetType(Manager))> _
Public Class Employee
    Public Name As String
End Class

Public Class Manager
    Inherits Employee
    Public Level As Integer
End Class


Public Class Run
    
    Public Shared Sub Main()
        Dim test As New
 Run()
        test.SerializeObject("IncludeExample.xml")
        test.DeserializeObject("IncludeExample.xml")
    End Sub    
    
    
    Public Sub SerializeObject(ByVal
 filename As String)
        Dim s As New XmlSerializer(GetType(Group))
        Dim writer As New
 StreamWriter(filename)
        Dim group As New
 Group()
        
        Dim manager As New
 Manager()
        Dim emp1 As New
 Employee()
        Dim emp2 As New
 Employee()
        
        manager.Name = "Zeus"
        manager.Level = 2
        
        emp1.Name = "Ares"
        
        emp2.Name = "Artemis"
        
        Dim emps() As Employee = {manager,
 emp1, emp2}
        group.Employees = emps
        
        s.Serialize(writer, group)
        writer.Close()
    End Sub    
    
    Public Sub DeserializeObject(ByVal
 filename As String)
        Dim fs As New FileStream(filename,
 FileMode.Open)
        Dim x As New XmlSerializer(GetType(Group))
        Dim g As Group = CType(x.Deserialize(fs),
 Group)
        Console.Write("Members:")
        
        Dim e As Employee
        For Each e In  g.Employees
            Console.WriteLine(ControlChars.Tab + e.Name)
        Next e
    End Sub
End Class

C#

using System;
using System.IO;
using System.Xml.Serialization;

public class Group
{   
   public Employee[] Employees;
}   

// Instruct the XmlSerializer to accept Manager types as well.
[XmlInclude(typeof(Manager))]
public class Employee
{
   public string Name;
}

public class Manager:Employee
{
   public int Level;
}


public class Run
{
   public static void Main()
   {
      Run test = new Run();
      test.SerializeObject("IncludeExample.xml");
      test.DeserializeObject("IncludeExample.xml");
   }


   public void SerializeObject(string
 filename)
   {
      XmlSerializer s = new XmlSerializer(typeof(Group));
      TextWriter writer = new StreamWriter(filename);
      Group group = new Group();
      
      Manager manager = new Manager();
      Employee emp1 = new Employee();
      Employee emp2 = new Employee();

      manager.Name = "Zeus";
      manager.Level = 2;

      emp1.Name = "Ares";

      emp2.Name = "Artemis";

      Employee [] emps = new Employee[3]{manager, emp1, emp2};
      group.Employees = emps;

      s.Serialize(writer, group);
      writer.Close();
   }

   public void DeserializeObject(string
 filename)
   {
      FileStream fs = new FileStream(filename, FileMode.Open);
      XmlSerializer x = new XmlSerializer(typeof(Group));
      Group g = (Group) x.Deserialize(fs);
      Console.Write("Members:");
      
      foreach(Employee e in g.Employees) 
      {
         Console.WriteLine("\t" + e.Name);
      }
   }
}

C++

#using <System.Xml.dll>
#using <System.dll>

using namespace System;
using namespace System::IO;
using namespace System::Xml::Serialization;

ref class Employee;

ref class Manager;

ref class Group
{
public:
   array<Employee^>^Employees;
};

// Instruct the XmlSerializer to accept Manager types as well.
[XmlInclude(Manager::typeid)]
public ref class Employee
{
public:
   String^ Name;
};

public ref class Manager: public
 Employee
{
public:
   int Level;
};

void SerializeObject( String^ filename )
{
   XmlSerializer^ s = gcnew XmlSerializer( Group::typeid );
   TextWriter^ writer = gcnew StreamWriter( filename );
   Group^ group = gcnew Group;
   Manager^ manager = gcnew Manager;
   Employee^ emp1 = gcnew Employee;
   Employee^ emp2 = gcnew Employee;
   manager->Name = "Zeus";
   manager->Level = 2;
   emp1->Name = "Ares";
   emp2->Name = "Artemis";
   array<Employee^>^emps = {manager,emp1,emp2};
   group->Employees = emps;
   s->Serialize( writer, group );
   writer->Close();
}

void DeserializeObject( String^ filename )
{
   FileStream^ fs = gcnew FileStream( filename,FileMode::Open );
   XmlSerializer^ x = gcnew XmlSerializer( Group::typeid );
   Group^ g = dynamic_cast<Group^>(x->Deserialize( fs ));
   Console::Write( "Members:" );
   System::Collections::IEnumerator^ myEnum = g->Employees->GetEnumerator();
   while ( myEnum->MoveNext() )
   {
      Employee^ e = safe_cast<Employee^>(myEnum->Current);
      Console::WriteLine( "\t{0}", e->Name );
   }
}

int main()
{
   SerializeObject( "IncludeExample.xml" );
   DeserializeObject( "IncludeExample.xml" );
}

J#

import System.*;
import System.IO.*;
import System.Xml.Serialization.*;

public class Group
{
    public Employee employees[];
} //Group

// Instruct the XmlSerializer to accept Manager types as well.
/** @attribute XmlInclude(Manager.class)
 */
public class Employee
{
    public String name;
} //Employee

public class Manager extends Employee
{
    public int level;
} //Manager

public class Run
{
    public static void main(String[]
 args)
    {
        Run test = new Run();
        test.SerializeObject("IncludeExample.xml");
        test.DeserializeObject("IncludeExample.xml");
    } //main

    public void SerializeObject(String fileName)
    {
        XmlSerializer s = new XmlSerializer(Group.class.ToType());
        TextWriter writer = new StreamWriter(fileName);
        Group group = new Group();
        
        Manager manager = new Manager();
        Employee emp1 = new Employee();
        Employee emp2 = new Employee();

        manager.name = "Zeus";
        manager.level = 2;
        emp1.name = "Ares";
        emp2.name = "Artemis";

        Employee emps[] = new Employee[] { manager, emp1, emp2
 };

        group.employees = emps;
        s.Serialize(writer, group);
        writer.Close();
    } //SerializeObject

    public void DeserializeObject(String fileName)
    {
        FileStream fs = new FileStream(fileName, FileMode.Open);
        XmlSerializer x = new XmlSerializer(Group.class.ToType());
        Group g = (Group)x.Deserialize(fs);

        Console.Write("Members:");
        for (int iCtr = 0; iCtr < g.employees.length;
 iCtr++) {
            Employee e = g.employees[iCtr];
            Console.WriteLine("\t" + e.name);
        }
    } //DeserializeObject
} //Run

プラットフォーム

Windows 98, Windows 2000 SP4, Windows CE, Windows Millennium Edition, Windows Mobile for Pocket PC, Windows Mobile for Smartphone, Windows Server 2003, Windows XP Media Center Edition, Windows XP Professional x64 Edition, Windows XP SP2, Windows XP Starter Edition

開発プラットフォームの中には、.NET Framework によってサポートされていないバージョンがあります。サポートされているバージョンについては、「システム要件」を参照してください。

バージョン情報

.NET Framework
サポート対象 : 2.0、1.1、1.0
.NET Compact Framework
サポート対象 : 2.0、1.0

参照

関連項目
XmlIncludeAttribute クラス
XmlIncludeAttribute メンバ
System.Xml.Serialization 名前空間