I have created a app using VFP 6.0 (SP3) and my customer is running it on a Pentium 75. The app starts slowly and if the user double clicks the shortcut icon a second time, they get an error message. The message is "access denied". I understand that a table was open exclusivly by the first execution and that's why I'm getting the error message. This is a standalone app and they should only start the app one time.

Is there a way to prevent the user from starting the app a second time?

Thanks
Mark Parker
MAP Software, Inc.