Getting the source directory of a Bash script from

Question

How do I get the path of the directory in which a Bash script is located, inside that script?

For instance, let's say I want to use a Bash script as a launcher for another application. I want to change the working directory to the one where the Bash script is located, so I can operate on the files in that directory, like so:

$ ./application

Answer 1

Short answer:

`dirname $0`

or (preferably):

$(dirname "$0")

Answer 2

$_ is worth mentioning as an alternative to $0. If you're running a script from bash, the accepted answer can be shortened to:

DIR="$( dirname "$_" )"

Note that this has to be the first statement in your script.

Answer 3

Use dirname "$0":

#!/bin/bash
echo "The script you are running has basename `basename "$0"`, dirname `dirname "$0"`"
echo "The present working directory is `pwd`"

using pwd alone will not work if you are not running the script from the directory it is contained in.

[matt@server1 ~]$ pwd
/home/matt
[matt@server1 ~]$ ./test2.sh
The script you are running has basename test2.sh, dirname .
The present working directory is /home/matt
[matt@server1 ~]$ cd /tmp
[matt@server1 tmp]$ ~/test2.sh
The script you are running has basename test2.sh, dirname /home/matt
The present working directory is /tmp

Answer 4

I've compared many of the answers given, and come up with some more compact solutions. These seem to handle all of the crazy edge cases that arise from your favorite combination of:

• Absolute paths or relative paths
• File and directory soft links
• Invocation as script, bash script, bash -c script, source script, or . script
• Spaces, tabs, newlines, unicode, etc. in directories and/or filename
• Filenames beginning with a hyphen

If you're running from Linux, it seems that using the proc handle is the best solution to locate the fully resolved source of the currently running script (in an interactive session, the link points to the respective /dev/pts/X):

resolved="$(readlink /proc/$$/fd/255 && echo X)" && resolved="${resolved%$'\nX'}"

This has a small bit of ugliness to it, but the fix is compact and easy to understand. We aren't using bash primitives only, but I'm okay with that because readlink simplifies the task considerably. The echo X adds an X to the end of the variable string so that any trailing whitespace in the filename doesn't get eaten, and the parameter substitution ${VAR%X} at the end of the line gets rid of the X. Because readlink adds a newline of its own (which would normally be eaten in the command substitution if not for our previous trickery), we have to get rid of that, too. This is most easily accomplished using the $'' quoting scheme, which lets us use escape sequences such as \n to represent newlines (this is also how you can easily make deviously named directories and files).

The above should cover your needs for locating the currently running script on Linux, but if you don't have the proc filesystem at your disposal, or if you're trying to locate the fully resolved path of some other file, then maybe you'll find the below code helpful. It's only a slight modification from the above one-liner. If you're playing around with strange directory/filenames, checking the output with both ls and readlink is informative, as ls will output "simplified" paths, substituting ? for things like newlines.

absolute_path=$(readlink -e -- "${BASH_SOURCE[0]}" && echo x) && absolute_path=${absolute_path%?x}
dir=$(dirname -- "$absolute_path" && echo x) && dir=${dir%?x}
file=$(basename -- "$absolute_path" && echo x) && file=${file%?x}

ls -l -- "$dir/$file"
printf '$absolute_path: "%s"\n' "$absolute_path"

Answer 5

pushd . > /dev/null
SCRIPT_PATH="${BASH_SOURCE[0]}"
if ([ -h "${SCRIPT_PATH}" ]); then
  while([ -h "${SCRIPT_PATH}" ]); do cd `dirname "$SCRIPT_PATH"`; 
  SCRIPT_PATH=`readlink "${SCRIPT_PATH}"`; done
fi
cd `dirname ${SCRIPT_PATH}` > /dev/null
SCRIPT_PATH=`pwd`;
popd  > /dev/null

Works for all versions,including

• when called via multple depth soft link,
• when the file it
• when script called by command "source" aka . (dot) operator.
• when arg $0 is modified from caller.
• "./script"
• "/full/path/to/script"
• "/some/path/../../another/path/script"
• "./some/folder/script"

Alternatively, if the bash script itself is a relative symlink you want to follow it and return the full path of the linked-to script:

pushd . > /dev/null
SCRIPT_PATH="${BASH_SOURCE[0]}";
if ([ -h "${SCRIPT_PATH}" ]) then
  while([ -h "${SCRIPT_PATH}" ]) do cd `dirname "$SCRIPT_PATH"`; SCRIPT_PATH=`readlink "${SCRIPT_PATH}"`; done
fi
cd `dirname ${SCRIPT_PATH}` > /dev/null
SCRIPT_PATH=`pwd`;
popd  > /dev/null

SCRIPT_PATH is given in full path, no matter how it is called.
Just make sure you locate this at start of the script.

This comment and code Copyleft, selectable license under the GPL2.0 or later or CC-SA 3.0 (CreativeCommons Share Alike) or later. (c) 2008. All rights reserved. No warranty of any kind. You have been warned.
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Answer 6

I would use something like this:

# retrieve the full pathname of the called script
scriptPath=$(which $0)

# check whether the path is a link or not
if [ -L $scriptPath ]; then

    # it is a link then retrieve the target path and get the directory name
    sourceDir=$(dirname $(readlink -f $scriptPath))

else

    # otherwise just get the directory name of the script path
    sourceDir=$(dirname $scriptPath)

fi