Get container directory more easily

Question

I have a bash script here:

$GOPATH/
     src/
     build.sh

and in build.sh I have:

export GOPATH="$(cd $(dirname "$BASH_SOURCE") && pwd)"

is there a shorter way to get the containing dir of build.sh?

score 2 · Answer 1 · answered Oct 19 '18 at 07:14

2

To get the directory where the script is located, use this:

readlink -f $(dirname $0)

As said in the bash man page, $0 is set to the name of the file.

readlink -f gets the absolute path of that directory.

answered Oct 19 '18 at 07:14

oliv

2,586
9
14

Unfortunately `readlink -f` is not very x-platform, it doesn't work on OSX sadly – Alexander Mills Oct 19 '18 at 17:32
`realpath` would also work, but I don't think realpath is available out of the box on OSX either, lame lol – Alexander Mills Oct 19 '18 at 17:34
Will not work for subscript executed inside original script – Nam G VU Nov 29 '22 at 17:45

score 0 · Answer 2 · answered Oct 23 '18 at 01:14

Extending the BASH_SOURCE idea, check to see if it contains an absolute path; if so, use it directly, otherwise prepend PWD. Afterwards, strip off the trailing slash and anything following it, leaving just the containing directory:

case ${BASH_SOURCE[0]} in
  ( /* )
        p=${BASH_SOURCE[0]}
        ;;
  ( * )
        p=${PWD}/${BASH_SOURCE[0]}
        ;;
esac

p=${p%/*}
printf "%s\n" "$p"

Get container directory more easily

2 Answers2