Autoexec.bat file scripting help required.

[myelin]

Hello guys

1. I want to know is there any way i can execute a script right at the beginning before Windows98SE boot process starts?

I want to run Puppy Linux, which i can be run by running go.bat (which i made, because it was easy), but i have to go into into its directory and then execute go.bat from DOS prompt everytime i want to run Linux, so is there any way to make it run before windows; if i press yes, and if i press no then it boots to Windows. Any lines which i can add to grub? or is there something else?

Thanks.

[jaclaz]

Well, of course you can do it BOTH ways, i.e. add an entry in Grub4dos for the two os and boot from it, or make some changes in autoexec.bat to have a choice when 98 boots.

You need either CHOICE.COM (included in DOS/Win9x) or CHOIX.COM, see here:

http://www.msfn.org/board/index.php?showtopic=81722

In your case will be something like this:

ECHO   Which OS do you want to load?
ECHO.
ECHO	1. Puppy Linux
ECHO.
ECHO	2. Windows 98
ECHO.
ECHO.
CHOICE /C:12 /N Please choose a menu option (1 or 2): 
IF ERRORLEVEL == 2 GOTO :SKIP
IF ERRORLEVEL == 1 GOTO :PUPPY

:PUPPY
CD "C:\your directory"
go.bat
EXIT

:SKIP
::put your normal autoexec.bat lines here:
.....
.....
.....
win.com

You will need to edit MSDOS.SYS as to start Win9x with no GUI:

http://www.annoyances.org/exec/show/article07-021

BootGUI=0

Or you can use the CONFIG.SYS method:

http://tldp.org/HOWTO/text/Loadlin+Win95-98-ME

(just think as your "go.bat" as the "linux.bat" referenced in the howto)

jaclaz

Edited March 4, 2007 by jaclaz

[myelin]

The first method was very simple and easy and it worked like a charm. Thanks for writing the whole script for me.

[jaclaz]

You are welcome.

jaclaz