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

If you have /proc:

$(cat /proc/$PPID/comm)

Answer 2

Based on @user3100381's answer, here's a much simpler command to get the same thing which I believe should be fairly portable:

PARENT_COMMAND=$(ps -o comm= $PPID)

Replace comm= with args= to get the full command line (command + arguments). The = alone is used to suppress the headers.

See: http://pubs.opengroup.org/onlinepubs/009604499/utilities/ps.html

Answer 3

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