On Thu, Jul 24, 2008 at 14:58, Frank Terbeck ft@grml.org wrote:
- Don't touch the current zshrc at all. It works fine and should be left intact unmodified.
We'll need to stop the normal zshrc from sourcing, so something like:
If there _is_ no normal zshrc, zsh-newuser-install will run. As we can easily replace this function, ZSH does the handling of this for free. The default zsh could be loaded the second a user presses enter on the little welcome screen menu grml displays on tty1 and everything set up to use this config, as well. Said menu would run from zsh-newuser-install then, btw.
- Put *every* function file, that does not collide with the ones that are already in grml's configuration into /etc/zsh/functions.d; There's no reason from keeping non-colliding function to not be in the standard config.
Unless, the function does only have value in connection with the configuration in question.
As I autoload every one of my functions in my zshrc via globbing, I would want to keep softlinks for the functions moved into main, but yes, that is a good idea.
- add /etc/zsh/rcs/richih.d/ at the *beginning* be $fpath so that colliding functions are not taken from the default dir, but from the dir that is part of the config, that is in action.
[[ -n $ZSHRC_ALT ]] && . /etc/zsh/rcs/$ZSHRC_ALT.d
in all zshrcs would mean you can fork and otherwise mangle them without needing to touch anything in the actual file.
Do NOT source /etc/zsh/zshrc in that case.
This should exist and overwrite the normal zsh-newuser-install.
Do NOT source ~/.zshrc in that case (set $ZDOTDIR to ~/.zshrc.$ZSHRC_ALT - that would make ~/.zshrc.$ZSHRC_ALT/.zshrc to be sourced; standard zsh mechanism.)We should touch a .zshrc in there if a given configuration does not provide one in order to avoid the newuser-module-dialog.
I would rather have a ~/.zshrc that simply uses $ZDOTDIR to source everything. Solves two problems in one go.
Do the handling of $ZSHRC_ALT, $ZDOTDIR and $fpath in the global zshenv (/etc/zsh/zshenv) in a
if [[ -o interactive ]] ; then fiBlock;
Would also be solved by the above, so make that three.
Richard