从C＃上的Windows shell上下文菜单中获取多个文件（参数）

Question

I am writing a C# application and it takes files as argument, I added it to shell context menu with code listed below;

if (((CheckBox)sender).CheckState == CheckState.Checked)
            {
                RegistryKey key = Registry.CurrentUser.OpenSubKey("Software\\Classes\\*\\shell\\" + KEY_NAME + "\\command");

                if (key == null)
                {
                    key = Registry.CurrentUser.CreateSubKey("Software\\Classes\\*\\shell\\" + KEY_NAME + "\\command");
                    key.SetValue("", Application.ExecutablePath + " \"%1\"");
                }
            }
            else if (((CheckBox)sender).CheckState == CheckState.Unchecked)
            {
                RegistryKey key = Registry.CurrentUser.OpenSubKey("Software\\Classes\\*\\shell\\" + KEY_NAME);

                if (key != null)
                {
                    Registry.CurrentUser.DeleteSubKeyTree("Software\\Classes\\*\\shell\\" + KEY_NAME);
                }

It is working good, but if I select multiple files, multiple instances of application running. for example if I select 5 files 5 application is opening, how can I fix this?

Answer 1

Detect if an instance of your application is already running on startup.

If it does, send the command line arguments to the running instance and exit the new instance.