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On of our Debian-servers won't boot into the newest installed kernel, and during troubleshooting I came across this:
root@server:~# grep GRUB_DEFAULT /etc/default/grub
GRUB_DEFAULT="1>2"
I haven't seen this syntax before. Does anyone know what this means?
From GNU GRUB Manual 2.02: default:
If the entry is in a submenu, then it must be identified using the number, title, or id of each of the submenus starting from the top level, followed by the number, title, or id of the menu entry itself, with each element separated by ‘>’
Meaning that
GRUB_DEFAULT="1>2"
Represents the following logic:
If entry 1 is a submenu, open it and select entry 2 from there.
Which is most likely the advanced submenu leading to a fallback/backup/recovery kernel.
2 in above example, what it points to, from command line, by looking at the grub menu-options configuration - sudo grub-mkconfig | grep -iE "menuentry 'Ubuntu, with Linux" | awk '{print i++ " : "$1, $2, $3, $4, $5, $6, $7}' [4 more details : askubuntu.com/a/1393019/350255]
Jul 15, 2022 at 5:07
Take note that for syntax like "1>2", the index starts from 0. So if the 3rd item is a submenu inside which the 2nd item is the default you want to boot, then it should be "2>1".
