bash - Run a script in the same directory as the current script

Question

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:

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)
2. 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?

Answer 1

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

so

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.