Run a script in the same directory as the current script

Bash Problem Overview

I have two Bash scripts in the same folder (saved somewhere by the user who downloads the entire repository):

script.sh is run by the user
helper.sh is required and run by script.sh

The two scripts should be in the same directory. I need the first script to call the second one, but there are two problems:

Knowing the current working directory is useless to me, because I don't know how the user is executing the first script (could be with /usr/bin/script.sh, with ./script.sh, or it could be with ../Downloads/repo/scr/script.sh)
The script script.sh will be changing to a different directory before calling helper.sh.

I can definitely hack together Bash that does this by storing the current directory in a variable, but that code seems needlessly complicated for what I imagine is a very common and simple task.

Is there a standard way to reliably call helper.sh from within script.sh? And will work in any Bash-supported operating system?

Bash Solutions

Solution 1 - Bash

Since $0 holds the full path of the script that is running, you can use dirname against it to get the path of the script:

#!/bin/bash

script_name=$0
script_full_path=$(dirname "$0")

echo "script_name: $script_name"
echo "full path: $script_full_path"

so if you for example store it in /tmp/a.sh then you will see an output like:

$ /tmp/a.sh
script_name: /tmp/a.sh
full path: /tmp

> 1. Knowing the current working directory is useless to me, because I don't know how the user is executing the first script (could be with > /usr/bin/script.sh, with ./script.sh, or it could be with > ../Downloads/repo/scr/script.sh)

Using dirname "$0" will allow you to keep track of the original path.

> 2. The script script.sh will be changing to a different directory before calling helper.sh.

Again, since you have the path in $0 you can cd back to it.

Solution 2 - Bash

$0 is considered unsafe and fragile by many devs. I have found another solution, it is safe for a chain of bash scripts and source.

If a.sh needs to execute b.sh (located in the same folder) using a child bash process:

#!/bin/bash
__dir="$(cd "$(dirname "${BASH_SOURCE[0]}")" && pwd)"
bash ${__dir}/b.sh

If a.sh needs to execute b.sh (located in the same folder) using the same bash process:

#!/bin/bash
__dir="$(cd "$(dirname "${BASH_SOURCE[0]}")" && pwd)"
source ${__dir}/b.sh

Solution 3 - Bash

> Is there a standard way to reliably call helper.sh from within > script.sh? And will work in any Bash-supported operating system?

In most cases, when helper.sh is in the same directory as script.sh, you can use in script.sh command:

. ${0%/*}/helper.sh

Explanation:
$0 stores the name of your process (in most cases it's the full path to your script).
${parameter%word} removes suffix pattern word from variable $parameter (in the command above it removes file name /* from the full path stored in variable $0).

If for some reasons (described in other answers) you don't want to use $0, you can use $BASH_SOURCE instead:

. ${BASH_SOURCE%/*}/helper.sh

And if you want, you can use source instead of .:

source ${BASH_SOURCE%/*}/helper.sh

As for me, it's the easiest way to achieve your goal.

Solution 4 - Bash

You can use the $0 variable. But this won't be as simple as getting current script name:

d=${0%/*}
[ x"$d" = x"$0" ] && d=.   # portable variant, use [[ ...]] in bash freely
readonly d                 # optional, for additional safety

# ... later on
. "$d"/helper.sh

This works well in set -e case as well.

Content Type	Original Author	Original Content on Stackoverflow
Question	IQAndreas	View Question on Stackoverflow
Solution 1 - Bash	fedorqui	View Answer on Stackoverflow
Solution 2 - Bash	Eugen	View Answer on Stackoverflow
Solution 3 - Bash	Ivan Olshansky	View Answer on Stackoverflow
Solution 4 - Bash	Vadim Zhukov	View Answer on Stackoverflow