Eclipse插件，如何使用弹出窗口操作向导？

Question

I wrote a wizard, now I want use a popup menu to launch the wizard. I make a action like this:

/**
 * @see IActionDelegate#run(IAction)
 */
public void run(IAction action) {

    MyWizard wizard = new MyWizard ();
    wizard.init(PlatformUI.getWorkbench(), new StructuredSelection());

    WizardDialog dialog = new WizardDialog(shell, wizard);
    dialog.create();
    dialog.getShell().setText(wizard.getWindowTitle());
    int result = dialog.open();
}

but it didn't work.

How can I do that ?

Answer 1

Actions are legacy. Did you try Commands? See here for an example