On Tue, 28 Jan 2014 15:13:17 -0600 "David C. Rankin" <drankinatty@xxxxxxxxxxxxxxxxxx> wrote: > All, > > The i686 archroot on the x86_64 box worked perfectly building i686 > TDE. However, as I tweak packages, I need to 'uninstall' a package > from the rw-layer without cleaning the entire archroot. Since this is > i686 on a x86_64 box how do I properly chroot the archroot so that I > can run pacman -R package name? > > Can I somehow: > > mount --bind some_dev $CHROOT/dev > mount -t proc none $CHROOT/proc > mount -t sysfs none $CHROOT/sys > cd $CHROOT > some_chroot_cmd > > So that the i686 system will be functional enough to remove a > package? I have seen the Arch64_FAQ page discussing the linux32 > wrapper working with i686 chroots created by installing with i686 ISO > "quickinstall", but this setup is just an i686 archroot. (will it > work?) > > I want to avoid cleaning the archroot or deleting and recreating a > new one just to test minor changes to package content adjustments due > the setup requiring several hundred dependencies and packages to > rebuild. > > What say the experts? > If your host is on Arch, you can do: pacman -R <pkgname> --arch i686 --root $CHROOT or install the Arch Install Scripts and chroot: pacman -S arch-install-scripts arch-chroot $CHROOT
