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From: | Joachim Mammele |
Subject: | Re: select menu item depending on frontkey |
Date: | Mon, 23 Apr 2012 15:28:15 +0200 |
User-agent: | Mozilla/5.0 (Windows NT 6.1; WOW64; rv:11.0) Gecko/20120327 Thunderbird/11.0.1 |
Okay, I modified the code. But still getting some error messages on the beginning of starting grub (and later on: "error: file not found, you need to load the kernel first") #set keybits = "32"; grub_outb(32, 0x20A); grub_outw(-16832, 0x20C); bits9 = grub_inb(0x209); echo "bits9= " echo ${bits9} if [${bits9} = 64]; then set default="1" elif [${bits9} = 32]; then set default="2" elif [${bits9} = 16]; then set default="3" else set default="0" fi Am 23.04.2012 14:56, schrieb Vladimir 'φ-coder/phcoder' Serbinenko: On 23.04.2012 14:54, Joachim Mammele wrote:Hi, accrding to Vladimirs hints I adapted my grub-configuration. What I changed in /boot/grub/grub.cfg is the following: (I'm aware that editing grub.cfg isn't the best option as it gets rewritten on executing the config-scripts) (the whole file can be found here http://pastebin.com/K57RHjwn) ### BEGIN /etc/grub.d/00_header ### if [ -s $prefix/grubenv ]; then load_env fi #unsigned char bits9, keybits; set keybits = "32"; grub_outb(keybits, 0x20A); grub_outw(-16832, 0x20C); set bits9 = grub_inb(0x209); echo "bits9= " echo bits9 if (bits9 == 64) set default="1" if (bits9 == 32) set default="2" if (bits9 == 16) set default="3" else set default="0" On starting grub I see some error log-messages but before I can read them the grub-menu appears. I'm using ubuntu 10.04 (Kernel 2.6.32-40) Any help would be appreciated.grub.cfg is written in GRUB bash-like, not C. Modules are in C.Greetings Joachim _______________________________________________ Grub-devel mailing list address@hidden https://lists.gnu.org/mailman/listinfo/grub-devel -- Joachim Mammele Egelhaafstrasse 3 70565 Stuttgart Mobil: 0162/4603792 E-Mail: address@hidden |
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