Get the name of the caller script in bash script

user_1769515 0
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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

shell bash
9 Answers

Declare this:

PARENT_NAME=`ps -ocomm --no-header $PPID`

Thus you'll get a nice variable $PARENT_NAME that holds the parent's name.

user_39852050
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