Bash – How to find the first missing directory in a long path

bashdirectorylsshell

Imagine I have a path that doesn't exist:

$ ls /foo/bar/baz/hello/world
ls: cannot access /foo/bar/baz/hello/world: No such file or directory

But let's say /foo/bar does exist. Is there a quick way for me to determine that baz is the breaking point in the path?

I'm using Bash.

Best Answer

Given a canonical pathname, such as yours, this will work:

set -f --; IFS=/
for p in $pathname
do    [ -e "$*/$p" ] || break
      set -- "$@" "$p"
done; printf %s\\n "$*"

That prints through the last fully existing/accessible component of $pathname, and puts each of those separately into the arg array. The first nonexistent component is not printed, but it is saved in $p.

You might approach it oppositely:

until cd -- "$path" && cd -
do    case   $path  in
      (*[!/]/*)
              path="${path%/*}"
;;    (*)   ! break
      esac
done  2>/dev/null   && cd -

That will either return appropriately or will pare down $path as needed. It declines to attempt a change to /, but if successful will print both your current working directory and the directory to which it changes to stdout. Your current $PWD will be put in $OLDPWD as well.

Related Question