I want GRUB to hide and wait 2 seconds to boot straight to Windows 10 unless I hit Esc. I modified /etc/default/grub according to the docs:
'GRUB_TIMEOUT_STYLE'
If this option is unset or set to 'menu', then GRUB will display
the menu and then wait for the timeout set by 'GRUB_TIMEOUT' to
expire before booting the default entry. Pressing a key interrupts
the timeout.
If this option is set to 'countdown' or 'hidden', then, before
displaying the menu, GRUB will wait for the timeout set by
'GRUB_TIMEOUT' to expire. If <ESC> is pressed during that time, it
will display the menu and wait for input. If a hotkey associated
with a menu entry is pressed, it will boot the associated menu
entry immediately. If the timeout expires before either of these
happens, it will boot the default entry. In the 'countdown' case,
it will show a one-line indication of the remaining time.
And it's now looking like this:
GRUB_DEFAULT="0"
GRUB_TIMEOUT_STYLE="countdown"
GRUB_TIMEOUT="2"
Unfortunately, nothing changes and GRUB is still showing after running
sudo update-grub
I've also tried Grub Customizer and it has successfully changed the order list and the names of the options, but has not worked with the hidding.
I'm using Ubuntu 18.04.4 and Windows 10. Secure Boot has been enabled and disabled without any effect.
Anyone could lend a hand?
Thanks in advance, J.
EDIT: I have already seen the answer in here but If I set
GRUB_DISABLE_OS_PROBER=true
I cannot boot to Windows, given that it isn't recognized by GRUB.
Is there any solution to this?
