I need to launch a program automatically on bootup. I know I need to edit my autoexec.bat FILE, in wich I have but cannot seem to get the commands right to get my program to launch properly. These are the commands I use once the c: prompt appears.

C:>cd XYZ

C:/XYZ> XYZ


Thanks, Dan.

The commands you enter from the prompt are the same ones you put in your file:

cd xyz
xyz

In what way is your program not properly launching? Are there any error messages?

Need more info for a definitive answer. When the program "properly " launches it searches for a setup file and COULD not find it when I got it to do so upon bootup. Could my problem be with the echo command? I am using DOS 6.2 and have not modified the file any further other than adding this to the end.


c:>cd xyz
c:/xyz>xyz

Do I need the c: prompts? Sorry for the elementary question and thanks for the help.http://www.robvanderwoude.com/index.htmlWhat echo command? What startup file? There seems to be more to this than you're telling us. No, you don't need the prompts in the autoexec.bat (if fact it will fail if you do), but we can force the ISSUE:

c:
cd \xyz
xyz

Is this startup file in the xyz directory? Do you need to pass a reference to the startup file on the command LINE?

Get back to us. I have noticed that you typed the following commands.

c:>cd xyz
c:/xyz>xyz

C:/XYZ> (all of this is wrong to start with)
If this is just YOUR typo on this post, then ignore this reply.

It should be: C:\XYZ\XYZ.EXE or C:\XYZ\XYZ.COM or C:\XYZ\XYZ.BAT
Dont use the ">" at the end either. You actually dont need to type "C:\XYZ" as you already did CHANGE into the directory.