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

Based on @J.L.answer, with more in depth explanations (the only one command that works for me (linux)) :

cat /proc/$PPID/comm

gives you the name of the command of the parent pid

If you prefer the command with all options, then :

cat /proc/$PPID/cmdline

explanations :

• $PPID is defined by the shell, it's the pid of the parent processes
• in /proc/, you have some dirs with the pid of each process (linux). Then, if you cat /proc/$PPID/comm, you echo the command name of the PID

Check man proc