A simple (g)awk script that lists the grub menu from the command line. Typically used when you use grub-set-default or grub-reboot. Tested with Grub 2 on Ubuntu 22.04. Assumes grub.cfg is well formed.

grub-menulst.sh

gawk  'BEGIN {
  in_submenu=0
  menuindex=0
}

# Skip lines not menu related
!/.*menu.*{.*/ { next; }

# Do the work for each menu line
{

match( $0, /^[^'\'']*'\''([^'\'']*)'\''.*$/, menu_arr )

if ( $0 ~ /^submenu.*{.*/ ) {
  submenu=0
  in_submenu=1
}

if ( $0 ~ /^menuentry.*{.*/ ) {
  submenu=0
  in_submenu=0
}

if (submenu == 0 ) {
  print menuindex ": " menu_arr[1]
  menuindex++
} else {
  print "  " (menuindex - 1) ">" (submenu - 1) " " menu_arr[1]
}

if (in_submenu == 1 ) {
  submenu++
}

}' /boot/grub/grub.cfg

It seems my work was wasted. The following one-liner does the same job:

awk -F\' '$1=="menuentry " || $1=="submenu " {print i++ " : " $2}; /\tmenuentry / {print "\t" i-1">"j++ " : " $2};' /boot/grub/grub.cfg

PS! All of this is coming from https://askubuntu.com/q/599208/