Get the name of the directory where a script is executed

Question

I have some script, that uses files in directories around it. It uses

dirname $0

command. It should work from any directory where I run this script, but when I run a symbolic link that points to that script I get the path of symbolic link. So I get the output of dirname rather than the path of the script itself.

Any one know a way to get the path of where the script is run?

Answer 1

Get the real path to your script

if [ -L $0 ] ; then
    ME=$(readlink $0)
else
    ME=$0
fi
DIR=$(dirname $ME)

Answer 2

if you have realpath installed:

$(dirname $(realpath $0))

Answer 3

Unless I misunderstand you, the problem should be the same as the one in: How do you normalize a file path in Bash?

An option not mentioned there is the following python one-liner:

python2.6 -c "import os,sys; print os.path.realpath(sys.argv[1])" "$0"

Finally, remember to use double quotes around "$0".

Answer 4

A simpler solution:

dirname $(readlink -f $0)

Tested with on Ubuntu 14.04: which java returns /usr/bin/java, which is a symbolic link.

readlink -f `which java`

Returns /usr/lib/jvm/java-8-oracle/jre/bin/java

Finally,

dirname $(readlink -f `which java`)

Returns /usr/lib/jvm/java-8-oracle/jre/bin, which is the folder under which "java" is located.