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Here is a simple GRUB script that searches for the first "disk" containing a file at /EFI/BOOT/BOOTX64.EFI and then conditionally display a menu entry to boot from it:

Anxious Armadillo answered on August 12, 2023 Popularity 1/10 Helpfulness 1/10

answer Here is a simple GRUB script that searches for the first "disk" containing a file at /EFI/BOOT/BOOTX64.EFI and then conditionally display a menu entry to boot from it:

Here is a simple GRUB script that searches for the first "disk" containing a file at /EFI/BOOT/BOOTX64.EFI and then conditionally display a menu entry to boot from it:

Comment

# FROM LINUX

search --no-floppy --file --set=bootable_efi /EFI/BOOT/BOOTX64.EFI
if [ -n "${bootable_efi}" ]; then
  menuentry "EFI Bootable USB" --class unknown {
    set root="${bootable_efi}"
    chainloader /EFI/BOOT/BOOTX64.EFI
    boot
  }
fi


# FROM GRUB


grub> ls
grub> set root=(hd0,msdos5)
grub> chainloader /efi/boot/grubx64.efi
grub> boot

Popularity 1/10 Helpfulness 1/10 Language javascript

Source: stackoverflow.com

Tags: boot display file grub menu script typescript

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