851

I'm writing a Bash script. I need the current working directory to always be the directory that the script is located in.

The default behavior is that the current working directory in the script is that of the shell from which I run it, but I do not want this behavior.

5
  • Have you considered putting a wrapper script somewhere like /usr/bin to cd into the (hardcoded) proper directory and then execute your script? Commented Jul 28, 2010 at 1:02
  • 2
    Why do you need the directory of the script? There's probably a better way to solve the underlying problem. Commented Jul 28, 2010 at 3:59
  • 19
    I'd just like to point out that the behavior you call "obviously undesirable" is in fact entirely necessary -- if I run myscript path/to/file I expect the script to evaluate path/to/file relative to MY current directory, not whatever directory the script happens to be located in. Also, what would you have happen for a script run with ssh remotehost bash < ./myscript as the BASH FAQ mentions? Commented Jul 28, 2010 at 4:54
  • possible duplicate of Can a Bash script tell what directory it's stored in? Commented Apr 24, 2015 at 2:23
  • 4
    cd "${BASH_SOURCE%/*}" || exit Commented Jul 7, 2017 at 23:48

14 Answers 14

1026

TL;DR

#!/usr/bin/env bash
cd "$(dirname "$0")"

The explanation

How does this work and how does it deal with edge and corner cases?

  • You type a script invocation command into your interactive shell, which may or may not be bash.
  • That interactive shell tells the kernel to execute the script from the command, and the kernel then calls bash with some name which tells bash the script to run.
  • That bash then calls dirname with the bash argument which points to the script.
  • Finally, that bash then calls cd with the output of dirname as its argument.

script invocation command bash argument dirname argument cd argument
foo (found in $PATH at /path/to/foo) /path/to/foo /path/to/foo /path/to
bash foo foo foo .
/foo /foo /foo /
./foo ./foo ./foo .
"/pa th/to/foo" /pa th/to/foo /pa th/to/foo /pa th/to
"./pa th/to/foo" ./pa th/to/foo ./pa th/to/foo ./pa th/to
"../pa th/to/foo" ../pa th/to/foo ../pa th/to/foo ../pa th/to
"../../pa th/to/foo" ../../pa th/to/foo ../../pa th/to/foo ../../pa th/to
"pa th/to/foo" pa th/to/foo pa th/to/foo pa th/to
--help/foo --help/foo * N/A N/A
--help N/A ** N/A N/A

On symlinks

The cd command will follow symlinks if they are involved. A symlink usually exists to be followed, so in most situations following the symlink is the correct thing to do. Why would it be a problem for the code in the script to follow a symlink when it was absolutely fine to follow that very same symlink a few microseconds earlier when loading the script?

On command arguments starting with hyphens

Elaborating on the two cases of arguments starting with hyphens in above table (marked with * and **, respectively):

* There is only one case where the argument to the dirname could begin with a -, and that is the relative path case --help/foo. If the script is in a subdirectory named --help, the script execution will run bash --help/foo, and bash does not know that option --help/foo and will therefore abort with an error message. The script will never execute, so it does not matter what the cd "$(dirname "$0")" would have executed.

** Note that naming the script --help makes the shell not find the command when you are typing --help in the same directory. Alternatively, with $PATH containing the current directory, the script will be run as /path/to/--help or ./--help, always with something in front of the - character.

Unless bash introduces command line arguments with a parameter separated by a =, it is unlikely to impossible to pass a - argument to bash which contains a / later, and which is accepted by bash.

If you can rely on dirname accepting -- argument (bash builtin cd will certainly accept --), you can change the script snippet to

cd -- "$(dirname -- "$0")"

Please do comment if you can figure out a way to construct an argument beginning with - which can be sneaked past bash.

Nota bene

The TL;DR snippet also works with non-bash /bin/sh.

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20 Comments

dirname returns '.' when using bash under Windows. So, Paul's answer is better.
Also returns '.' in Mac OSX
It's worth noting that things can break if a symbolic link makes up part of $0. In your script you may expect, for example, ../../ to refer to the directory two levels above the script's location, but this isn't necessarily the case if symbolic links are in play.
If you called the script as ./script, . is the correct directory, and changing to . it will also end up in the very directory where script is located, i.e. in the current working directory.
If you run the script from the current directory like so bash script.sh, then the value of $0 is script.sh. The only way the cd command will "work" for you is because you don't care about failed commands. If you were to use set -o errexit (aka: set -e) to ensure that your script doesn't blow past failed commands, this would NOT work because cd script.sh is an error. The reliable [bash specific] way is cd "$(dirname ${BASH_SOURCE[0]})"
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402

The following also works:

cd "${0%/*}"

The syntax is thoroughly described in this StackOverflow answer.

12 Comments

Explanation how it works: stackoverflow.com/questions/6393551/…
Don't forget to enclose in quotes if the path contains whitespace. i.e. cd "${0%/*}"
This fails if the script is called with bash script.sh -- $0 will just be the name of the file
neither it works if a script is in the root directory, because "${0%/*}" would expand to empty string
I would say this answer is too succinct. You hardly learn anything about the syntax. Some description about how to read this would be helpful.
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257

Try the following simple one-liners:


For all UNIX/OSX/Linux

dir="$(cd -P -- "$(dirname -- "$0")" && pwd -P)"

Bash

dir="$(cd -P -- "$(dirname -- "${BASH_SOURCE[0]}")" && pwd -P)"

Note: A double dash (--) is used in commands to signify the end of command options, so files containing dashes or other special characters won't break the command.

Note: In Bash, use ${BASH_SOURCE[0]} in favor of $0, otherwise the path can break when sourcing it (source/.).


*For Linux, Mac and other BSD:

cd "$(dirname "$(realpath -- "$0")")";

Note: realpath should be installed in the most popular Linux distribution by default (like Ubuntu), but in some it can be missing, so you have to install it.

Note: If you're using Bash, use ${BASH_SOURCE[0]} in favor of $0, otherwise the path can break when sourcing it (source/.).

Otherwise you could try something like that (it will use the first existing tool):

cd "$(dirname "$(readlink -f -- "$0" || realpath -- "$0")")"

For Linux specific:

cd "$(dirname "$(readlink -f -- "$0")")"

*Using GNU readlink on BSD/Mac:

cd "$(dirname "$(greadlink -f -- "$0")")"

Note: You need to have coreutils installed (e.g. 1. Install Homebrew, 2. brew install coreutils).


In bash

In bash you can use Parameter Expansions to achieve that, like:

cd "${0%/*}"

but it doesn't work if the script is run from the same directory.

Alternatively you can define the following function in bash:

realpath () {
  [[ "$1" = /* ]] && echo "$1" || echo "$PWD/${1#./}"
}

This function takes 1 argument. If argument has already absolute path, print it as it is, otherwise print $PWD variable + filename argument (without ./ prefix).

or here is the version taken from Debian .bashrc file:

function realpath()
{
    f=$@
    if [ -d "$f" ]; then
        base=""
        dir="$f"
    else
        base="/$(basename -- "$f")"
        dir="$(dirname -- "$f")"
    fi
    dir="$(cd -- "$dir" && /bin/pwd)"
    echo "$dir$base"
}

Related:

See also:

How can I get the behavior of GNU's readlink -f on a Mac?

12 Comments

a much better answer than the popular ones because it resolves syslinks, and accounts for different OS's. Thanks!
A double dash (--) is used in commands to signify the end of command options, so files containing dashes or other special characters won't break the command. Try e.g. create the file via touch "-test" and touch -- -test, then remove the file via rm "-test" and rm -- -test, and see the difference.
I'd remove my upvote if I could: realpath is deprecated unix.stackexchange.com/questions/136494/…
@lsh It says that only Debian realpath package is deprecated, not GNU realpath. If you think it's not clear, you can suggest an edit.
Shouldn't all the examples like cd "$(dirname "$(realpath "$0")")" also use -- before the parameter, like cd "$(dirname -- "$(realpath -- "$0")")"?
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102
cd "$(dirname "${BASH_SOURCE[0]}")"

It's easy. It works.

5 Comments

This should be the accepted answer. Works on OSX and Linux Ubuntu.
This works great when the script B is sourced from other script A and you need to use paths relative to script B. Thank you.
This also works in Cygwin for those of us unfortunate enough to have to touch Windows.
Or cd "${BASH_SOURCE%/*}" || exit
Works on macOS Mojave
24

The accepted answer works well for scripts that have not been symlinked elsewhere, such as into $PATH.

#!/bin/bash
cd "$(dirname "$0")"

However if the script is run via a symlink,

ln -sv ~/project/script.sh ~/bin/; 
~/bin/script.sh

This will cd into the ~/bin/ directory and not the ~/project/ directory, which will probably break your script if the purpose of the cd is to include dependencies relative to ~/project/

The symlink safe answer is below:

#!/bin/bash
cd "$(dirname "$(readlink -f "${BASH_SOURCE[0]}")")"  # cd current directory

readlink -f is required to resolve the absolute path of the potentially symlinked file.

The quotes are required to support filepaths that could potentially contain whitespace (bad practice, but its not safe to assume this won't be the case)

Comments

18

There are a lot of correct answers in here, but one that tends to be more useful for me (making sure a script's relative paths remain predictable/work) is to use pushd/popd:

pushd "$(dirname ${BASH_SOURCE:0})"
trap popd EXIT

# ./xyz, etc...

This will push the source file's directory on to a navigation stack, thereby changing the working directory, but then, when the script exits (for whatever reason, including failure), the trap will run popd, restoring the current working directory before it was executed. If the script were to cd and then fail, your terminal could be left in an unpredictable state after the execution ends - the trap prevents this.

1 Comment

Now that doesn't work with '> /dev/null/ ' to make your popd silent anymore. Found this as a workaround: stackoverflow.com/a/25288289/4529404 Using trap is a cool idea, but this simple bash script is getting quite complicated quite fast.
12

This script seems to work for me:

#!/bin/bash
mypath=`realpath $0`
cd `dirname $mypath`
pwd

The pwd command line echoes the location of the script as the current working directory no matter where I run it from.

3 Comments

realpath is unlikely to be installed everywhere. Which may not matter, depending on the OP's situation.
My system doesn't have realpath but it does have readlink which seems to be similar.
Which dist doesn't have realpath installed by default? Ubuntu has it.
4

I take this and it works.

#!/bin/bash
cd "$(dirname "$0")"
CUR_DIR=$(pwd)

Comments

2

Get the real path to your script

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

(This is answer to the same my question here: Get the name of the directory where a script is executed)

Comments

1
cd "`dirname $(readlink -f ${0})`"

3 Comments

Can you explain your answer please ?
Although this code may help to solve the problem, it doesn't explain why and/or how it answers the question. Providing this additional context would significantly improve its long-term educational value. Please edit your answer to add explanation, including what limitations and assumptions apply.
${0} would give the file name of the script. readlink -f {0} would give that script's absolute path. dirname would extract the script located path from previous result.
1

Most answers either don't handle files which are symlinked via a relative path, aren't one-liners or don't handle BSD (Mac). A solution which does all three is:

HERE=$(cd "$(dirname "$BASH_SOURCE")"; cd -P "$(dirname "$(readlink "$BASH_SOURCE" || echo .)")"; pwd)

First, cd to bash's conception of the script's directory. Then readlink the file to see if it is a symlink (relative or otherwise), and if so, cd to that directory. If not, cd to the current directory (necessary to keep things a one-liner). Then echo the current directory via pwd.

You could add -- to the arguments of cd and readlink to avoid issues of directories named like options, but I don't bother for most purposes.

You can see the full explanation with illustrations here:

https://www.binaryphile.com/bash/2020/01/12/determining-the-location-of-your-script-in-bash.html

Comments

0

This solution worked for me:

cd "$(pwd)"

Comments

-10
echo $PWD

PWD is an environment variable.

1 Comment

This only gives the directory called from, not the directory where the script is.
-12

If you just need to print present working directory then you can follow this.

$ vim test

#!/bin/bash
pwd
:wq to save the test file.

Give execute permission:

chmod u+x test

Then execute the script by ./test then you can see the present working directory.

1 Comment

The question was how to ensure the script runs in its own directory, including if that is not the working directory. So this is not an answer. Anyway, why do this instead of just... running pwd? Seems like a lot of wasted keypresses to me.

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