Menu::MenuItemCollection::Contains Method (MenuItem^)
.NET Framework (current version)
Determines if the specified MenuItem is a member of the collection.
Assembly: System.Windows.Forms (in System.Windows.Forms.dll)
Parameters
- value
-
Type:
System.Windows.Forms::MenuItem^
The MenuItem to locate in the collection.
Return Value
Type: System::Booleantrue if the MenuItem is a member of the collection; otherwise, false.
In this example, you create a main menu, myMainMenu, with two MenuItem objects, File and Edit. The File menu has three submenu items, New, Open, and Exit. By using the Contains method, you check to see if the File menu collection contains the item Open. If the item exists, you display the result in a text box. This program requires that you have already created a Form named Form1.
public: void InitializeMenu() { // Create the MainMenu object. MainMenu^ myMainMenu = gcnew MainMenu; // Create the MenuItem objects. MenuItem^ fileMenu = gcnew MenuItem( "&File" ); MenuItem^ editMenu = gcnew MenuItem( "&Edit" ); MenuItem^ newFile = gcnew MenuItem( "&New" ); MenuItem^ openFile = gcnew MenuItem( "&Open" ); MenuItem^ exitProgram = gcnew MenuItem( "E&xit" ); // Add the MenuItem objects to myMainMenu. myMainMenu->MenuItems->Add( fileMenu ); myMainMenu->MenuItems->Add( editMenu ); // Add three submenus to the File menu. fileMenu->MenuItems->Add( newFile ); fileMenu->MenuItems->Add( openFile ); fileMenu->MenuItems->Add( exitProgram ); // Assign myMainMenu to the form. Menu = myMainMenu; // Check that the File menu contains the Open menu item. if ( fileMenu->MenuItems->Contains( openFile ) ) { MessageBox::Show( "The File menu contains 'Open' " + fileMenu->Text ); } }
.NET Framework
Available since 1.1
Available since 1.1
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