Excel添加事件在打开新工作表后停止触发

Question

我正在开发一个excel添加。当我的功能区加载时，我添加一个单元格上下文菜单，其中包含一个按钮。我将处理程序附加到按钮的单击事件以选择文件。当我运行excel时，它可以正常使用默认的第一张表。但是，如果我打开一个新工作表，单击时没有任何反应。为什么呢？

我的功能区加载功能：

    private void Ribbon1_Load(object sender, RibbonUIEventArgs e)
    {
        Helper.SetVar(Globals.ThisAddIn.Application, Globals.Factory);
    }

我有几个添加引用工具箱DLL。所以这是工具箱DLL的公共静态Helper类中的setVar函数：

    public static void SetVar(Excel.Application appSource, Microsoft.Office.Tools.Excel.ApplicationFactory FactorySource)
    {
        if(path=="")
        {
            string fullPath = typeof(Helper).Assembly.CodeBase;
            string theDirectory = Path.GetDirectoryName(fullPath);
            path = new Uri(theDirectory).LocalPath;
        }
        if (app == null)
            app = appSource;
        if (Factory == null)
            Factory = FactorySource;
        rightClickMenu();
    }

它调用rightClickMenu，即：

    public static void rightClickMenu()
    {
        bool val = false;
        removeRightClickMenu(ref val);
        Office.CommandBarPopup ctrl = (Office.CommandBarPopup)app.CommandBars["Cell"].Controls.Add(Type: Office.MsoControlType.msoControlPopup, Before: 1);
        ctrl.Caption = "Perso";
        Office.CommandBarButton btn = (Office.CommandBarButton)ctrl.Controls.Add(Type: Office.MsoControlType.msoControlButton);
        btn.Caption = "Sélectionner un fichier";
        btn.Click += new Office._CommandBarButtonEvents_ClickEventHandler(selectFile);
    }

点击处理程序是：

    public static void selectFile(Office.CommandBarButton ctrl, ref bool Cancel)
    {
        OpenFileDialog openFileDialog1 = new OpenFileDialog();
        openFileDialog1.InitialDirectory = "c:\\";
        openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*";
        openFileDialog1.FilterIndex = 2;
        openFileDialog1.RestoreDirectory = true;
        if (openFileDialog1.ShowDialog() == DialogResult.OK)
        {
            var filename = openFileDialog1.FileName;
            //var filename = openFileDialog1.SafeFileName;
            app.ActiveCell.Value = filename;
        }
    }