Linux / Unix: Shell Script Find Out In Which Directory Script File Resides

last updated in Categories , , , , , ,

I need to find out in which directory my bash script resides so that I can read config file called .backup .ignore .target. For example, if my script resides in >/home/foo/, I need to read /home/foo/.{backup,ignore,target} files. How do I find out the current directory location and shell script directory location in Bash running on Linux or Unix like operating systems?

You can use any one of the following method to find out the portion of pathname:


  1. basename command – Display filename portion of pathname.
  2. dirname command – Display directory portion of pathname.
  3. Bash parameter substitution.
  4. $0 expands to the name of the shell or shell script.

Examples: Shell script find out which directory the script file resides

The following example display directory path or portion of /home/nixcraft/scripts/

dirname /home/nixcraft/scripts/

Sample outputs:


The following line sets the shell variable i to /home/nixcraft/scripts:

i=`dirname /home/nixcraft/scripts/`
echo "$i"


i=$(dirname /home/nixcraft/scripts/
echo "$i"

In bash script use $0 instead of /home/nixcraft/scripts/

basename="$(dirname $script)"
echo "Script name $script resides in $basename directory."

Sample outputs:

Script name /tmp/ resides in /tmp directory.

Using bash shell ${var%pattern} syntax

To Remove from shortest rear (end) pattern use the following syntax:


For example:

echo "${x%/*}"
echo "$y"

An updated version of the above script:

# Purpose : Linux / Unix shell script find out which directory this script file resides
# Author : nixCraft <> under GPL v2.x+
# -------------------------------------------------------------------------------------
echo "Script name $script resides in $basename directory."
echo "Reading config file $config1 $config2 $config3, please wait..."

Run it as:
$ chmod +x /tmp/
$ /tmp/

Sample outputs:

Fig.01 Sample run from
Fig.01 Sample run from

A note about finding physical or real path

You may not get a real physical path and real path may be a symbolic link. To get physical path use realpath command. The realpath command uses the realpath() function to resolve all symbolic links, extra / characters and references to /./ and /../ in path. This is useful for shell scripting and security related applications.

Another recommended option is to use the readlink command to display value of a symbolic link or canonical file name:

# Purpose : Linux / Unix shell script find out which directory this script file resides
# Author : nixCraft <> under GPL v2.x+
# -------------------------------------------------------------------------------------
## Who am i? ##
## Get real path ##
_script="$(readlink -f ${BASH_SOURCE[0]})"
## Delete last component from $_script ##
_mydir="$(dirname $_script)"
## Delete /path/to/dir/ component from $_script ##
_myfile="$(basename $_script)"
echo "Script : $_script"
echo "Directory portion of $_script : $_mydir"
echo "Filename portion of $_script : $_myfile"

Save and close the file. Run it as follows:

cd /home/vivek/

Sample outputs:

Fig.02: Finding real path
Fig.02: Finding real path

See also
  • See man pages for more info – bash(1)


Posted by: Vivek Gite

The author is the creator of nixCraft and a seasoned sysadmin, DevOps engineer, and a trainer for the Linux operating system/Unix shell scripting. Get the latest tutorials on SysAdmin, Linux/Unix and open source topics via RSS/XML feed or weekly email newsletter.

11 comment

  1. i found aund used the function below. This should be included in the script itself.

    getScriptPath() {
        if [ -d ${0%/*} ]
            abspath=$(cd ${0%/*} && echo $PWD/${0##*/})
            # to get the path only - not the script name - add
            pathOnly=`dirname "$abspath"`
            progdir=`dirname $0`
            cd $progdir
        echo $pathOnly;

    Further down in the script I can use $(getScriptPath) as variable, such as

    mysql database <$(getScriptPath)/query.sql

    Hope this helps:

  2. Hello,

    The problem is that method won’t give your the absolute path if it is started with the relative one.
    For example, if my script is launched with ./ I won’t know from the script where am I really residing in the FS.

    So the response is :
    abspath=$(readlink -f $0)

    Then you can do any basename, dirname an other path manipulation on it.

    1. Chris,

      Can you tell me why would it be badly designed? Jugement without deep knowledge about the facts are useless.


      1. I feel if a script has to be edited when moved from directory to directory to operate correctly it is badly designed.

        1. If a script has to be moved from directory to directory, it is badly designed. Scripts should always be placed in a directory in your PATH variable, so they can be called from anywhere.

  3. readlink -f $0 won’t work on Mac OS X.
    A nice one-liner would be

    SCRIPT_PATH=$(cd $(dirname ${0}) && pwd)

    It doesn’t resolve the real path, though. For this you’ll need an OS check or something…

    SCRIPT_PATH=$(cd $(dirname ${0}) && pwd)
    echo $SCRIPT_PATH
    if [[ ${OSTYPE} = darwin* ]]; then
    	TEMP=`readlink ${SCRIPT_PATH}`
    	echo $TEMP
    	SCRIPT_PATH=`readlink -f ${SCRIPT_PATH}`

    Note that OS X’s readlink is not reliable.

    Still, have a question? Get help on our forum!