Compress working directory purely using bash regex - regex

Question

I want to know the best way to compress the current working directory so that only the last directory's full name is visible. Let me give an example:
$ echo $PWD
/Users/mac/workshop/project1/src
I want to be able to do bash regex replacement operations on it such that I can get ~/w/p/src
I can obtain the first part of getting the leading ~ by doing ${PWD/#$HOME/\~}
$ echo ${PWD/#$HOME/\~}
~/workshop/project1/src
What other regex operations can I do (is it possible to chain the regex operators?) so that I get the following
$ echo ${PWD/#$HOME/\~} ...
~/w/p/src
Note that I need to do only using bash i.e. no sed, awk, grep etc.
The intention for this is so that, I can set the PROMPT value based on bash i.e.
in my .bashrc, I want to:
export PROMPT=${PWD/#$HOME/\~}...

Answer 1

Do-able in just bash, but not as simple as you'd like:
$ squashPWD() {
local pwd parts part
IFS=/ read -ra parts <<< "${PWD/#$HOME/\~}"
for part in "${parts[#]:0:${#parts[#]}-1}"; do
pwd+="${part:0:1}/"
done
echo "$pwd${parts[-1]}"
}
$ pwd
/home/jackman/tmp/adir/foo
$ squashPWD
~/t/a/foo
$ cd /usr/local/share/doc/fish/
$ squashPWD
/u/l/s/d/fish
If you don't need bash:
squashPWD() { perl -pe 's/^$ENV{HOME}/~/; s{([^/])[^/]*(?=/)}{$1}g' <<<"$PWD"; }
Either way, your prompt can be something like:
PS1='\u#\h:$(squashPWD) \$ '

Answer 2

It doesn't need to be all bash, you can use a function in your bashrc or bash_profile and thus use sed or awk. You can put something like this in your bashrc:
short() {
local short_path=$(echo "$PWD" | sed -E 's!/(.)[^/]*!/\1!g')
local last_dir=${PWD##*/}
echo "${short_path::-1}${last_dir}" # remove last character (1st character of last directory, and just append the last directory)
}
PS1='$(short) '
Keep in mind I don't think I replace your $HOME directory with ~, but you know how to do that :)