Get the name of the caller script in bash script

Question

Let\'s assume I have 3 shell scripts:

script_1.sh

#!/bin/bash
./script_3.sh

script_2.sh

Answer 1

You can simply use the command below to avoid calling cut/awk/sed:

ps --no-headers -o command $PPID

If you only want the parent and none of the subsequent processes, you can use:

ps --no-headers -o command $PPID | cut -d' ' -f1