(Noobie question)
I have a functioning dual-boot PC, openSUSE 11.2 (default:)) and Windows/XP. GRUB manages the boot and menu.lst is:
# Modified by YaST2. Last modification on Thu Feb 11 18:16:48 EST 2010
# THIS FILE WILL BE PARTIALLY OVERWRITTEN by perl-Bootloader
# Configure custom boot parameters for updated kernels in /etc/sysconfig/bootloader
default 0
timeout 8
##YaST - generic_mbr
gfxmenu (hd0,5)/boot/message
##YaST - activate
###Don't change this comment - YaST2 identifier: Original name: linux###
title SUSE LINUX
root (hd0,5)
kernel /boot/vmlinuz root=/dev/disk/by-id/ata-HTS726060M9AT00_MRH423M4H4MRTB-part6 repair=1 resume=/dev/disk/by-id/ata-HTS726060M9AT00_MRH423M4H4MRTB-part5 splash=silent quiet showopts vga=0x317
initrd /boot/initrd
###Don't change this comment - YaST2 identifier: Original name: windows###
title Windows
rootnoverify (hd0,0)
chainloader +1
###Don't change this comment - YaST2 identifier: Original name: failsafe###
title Failsafe -- SUSE LINUX
root (hd0,5)
kernel /boot/vmlinuz root=/dev/disk/by-id/ata-HTS726060M9AT00_MRH423M4H4MRTB-part6 showopts apm=off noresume nosmp maxcpus=0 edd=off powersaved=off nohz=off highres=off processor.max_cstate=1 x11failsafe vga=0x317
initrd /boot/initrd
In prior experience (Windows-only dual-boots), I could set the timeout unusually high to avoid “default” booting. Now that GRUB is the boot manager, can I 1) similarly increase “timeout 8” to some large value, and 2) is it possible to either remove “Default” or set “Default” to none, such that a boot request will sit on the selection panel ?
(By way of an explanation: Windows updates often require a re-boot, and (occasionally) MS automatically queues said re-boot, with an automatic countdown. Further, some third-party products, such as Tallemu’s “Online Armor” and others programatically queue multiple boots (boot, do something, re-boot, resume action), and can get somewhat cranky when things do not go their way rotfl! )