c# WinForm Menu自动生成

1.将采用递归的方式生成窗体菜单,并且添加其事件响应

        Type t = typeof(frmMain);
        private void GenerateMenu(ToolStripItemCollection parentItem, List<SpinSystem.Domain.Main.Menu> menus,string paentId)
        {
            object o = this;

            if (menus == null) return;

            List<SpinSystem.Domain.Main.Menu> currentMenus = menus.Where(n => n.PARENT_ID == paentId).ToList();
            ToolStripMenuItem item = null;

            for (int i = 0; i < currentMenus.Count; i++)
            {
                if (!currentMenus[i].MENU_NAME.ToUpper().Equals("LINE"))
                {
                    item = new ToolStripMenuItem(currentMenus[i].MENU_DISPLAY);
                    item.Name = currentMenus[i].MENU_NAME;

                    MethodInfo info = t.GetMethod(currentMenus[i].LINK_FUNC);
                    if (info != null)
                    {
                        Delegate del = Delegate.CreateDelegate(typeof(EventHandler), o, info);
                        EventHandler myMethod = del as EventHandler;
                        item.Click += myMethod;
                    }

                    parentItem.Add(item);
                }
                else
                {
                    ToolStripSeparator line = new ToolStripSeparator();
                    parentItem.Add(line);
                    continue;
                }
                

                GenerateMenu(item.DropDownItems, menus, currentMenus[i].MENU_ID);
            }
        }

2.调用代码

 List<SpinSystem.Domain.Main.Menu> menus = services.GetMenuListByUserId(SpinSystem.Domain.Login.LoginUsercs.LoginUser.UserId);

 this.mnsMain.Items.Clear();
 GenerateMenu(this.mnsMain.Items,menus,"0");