Run a script in the same directory as the current script

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/bashscript_name=$0script_full_path=$(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.shscript_name: /tmp/a.shfull 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.

$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

Is there a standard way to reliably call helper.sh from withinscript.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.