I'd like to create a script that sets the $PATH to a set of directories chosen by me (thus ignoring the predefined paths, such as /usr/bin, /usr/local/bin and so on...) followed by launching ZSH.
This does the trick for BASH:
#!/bin/sh PATH="/Users/Thoht/Sites/djangostack/apps/django/bin" # exec /bin/bash --noprofile --norc
I get a new BASH session with the path being only what is defined above that I cancel with Ctrl-D when done. I'd like to recreate this script for ZSH but have had no success. I tried this:
#!/bin/sh PATH="/Users/Thoht/Sites/djangostack/apps/django/bin" # exec /bin/zsh -f
However the predefined paths such as /usr/bin and all that jazz remains which makes the script useless.
Preferably I'd like to have my .zshrc loaded and merely the path temporarily changed.
EDIT I forgot to add one thing - /etc/zshenv that is read first of all (or so I believe) contains this script:
# system-wide environment settings for zsh(1) if [ -x /usr/libexec/path_helper ]; then eval `/usr/libexec/path_helper -s` fi
EDIT 2 The first edit is irrelevant as I don't want to use the -f option. A more precise question is:
"How do I add paths to precede the .zshrc paths to $PATH in a shell script followed by launching ZSH normally (as in with it reading .zshrc)?"