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

#!/bin/bash ./script_3.sh 

the problem is that in script_3.sh I want to know the name of the caller script.

so that I can respond differently to each caller I support

please don't assume I'm asking about $0 cause $0 will echo script_3 every time no matter who is the caller

here is an example input with expected output

• ./script_1.sh should echo script_1

• ./script_2.sh should echo script_2

• ./script_3.sh should echo user_name or root or anything to distinguish between the 3 cases?

Is that possible? and if possible, how can it be done?

this is going to be added to a rm modified script... so when I call rm it do something and when git or any other CLI tool use rm it is not affected by the modification

Answer 1

In case you are sourceing instead of calling/executing the script there is no new process forked and thus the solutions with ps won't work reliably.

Use bash built-in caller in that case.

$ cat h.sh  #! /bin/bash  function warn_me() {      echo "$@"      caller  }  $ cat g.sh  #!/bin/bash  source h.sh  warn_me "Error: You didn't do something"  $ . g.sh  Error: You didn't do something 3  g.sh $ 

Source

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