如何让 grub 出现在组合键上,但让 windows 默认启动。
我正在运行 ubuntu 11.10 和 grub 2.0。
这是我当前的 /etc/default/grub
# If you change this file, run 'update-grub' afterwards to update
# /boot/grub/grub.cfg.
# For full documentation of the options in this file, see:
# info -f grub -n 'Simple configuration'
GRUB_DEFAULT=0
GRUB_HIDDEN_TIMEOUT=0
GRUB_HIDDEN_TIMEOUT_QUIET=true
GRUB_TIMEOUT=10
GRUB_DISTRIBUTOR=`lsb_release -i -s 2> /dev/null || echo Debian`
GRUB_CMDLINE_LINUX_DEFAULT="quiet splash"
GRUB_CMDLINE_LINUX=" quiet vga=769"
谢谢!
这是我的 /boot/grub/grub.cfghttp://pastebin.com/HbDBe8xz
答案1
我认为您可以使用中的配置变量来完成此操作。更新文件后/etc/default/grub
别忘了运行。update-grub
GRUB_DEFAULT= 3 #Assuming Windows is the third menu entry
GRUB_TIMEOUT=0
GRUB_HIDDEN_TIMEOUT=0
根据每个选项的文档:
`GRUB_DEFAULT'
The default menu entry. This may be a number, in which case it
identifies the Nth entry in the generated menu counted from zero,
or the title of a menu entry, or the special string `saved'.
Using the title may be useful if you want to set a menu entry as
the default even though there may be a variable number of entries
before it.
For example, if you have:
menuentry 'Example GNU/Linux distribution' --class gnu-linux {
...
}
then you can make this the default using:
GRUB_DEFAULT='Example GNU/Linux distribution'
If you set this to `saved', then the default menu entry will be
that saved by `GRUB_SAVEDEFAULT', `grub-set-default', or
`grub-reboot'.
The default is `0'.
`GRUB_TIMEOUT'
Boot the default entry this many seconds after the menu is
displayed, unless a key is pressed. The default is `5'. Set to
`0' to boot immediately without displaying the menu, or to `-1' to
wait indefinitely.
`GRUB_HIDDEN_TIMEOUT'
Wait this many seconds for a key to be pressed before displaying
the menu. If no key is pressed during that time, boot
immediately. Unset by default.