How to find out the initiator of a script in unix

Question

I am wondering if there is a way to find out how a file execution was initiated.

For example, consider the following files:

~/foo.sh

echo "Hello from foo.sh"
# print the name of the initiator/parent of this execution 

~/bar.sh

source ~/foo.sh

~/baz.sh

. ~/foo.sh

---

When I execute

sh ~/bar.sh or .~/bar.sh, ~/foo.sh should print ~/bar.sh

sh ~/baz.sh or .~/baz.sh, ~/foo.sh should print ~/baz.sh

---

I am trying to be generic but it could be specific to bash or zsh.

Answer 1

This is a bash solution (for scripts with #!/bin/bash as the first line, or run with bash script...).

Set up the example (two scripts, a.sh and b.sh):

cat >a.sh <<'x' && chmod a+x a.sh
#!/bin/bash
echo This is a.sh
source b.sh
echo End a.sh
x

cat >b.sh <<'x' && chmod a+x b.sh
#!/bin/bash
echo This is b.sh
echo "BASH_SOURCE=(${BASH_SOURCE[@]}) and we are '${BASH_SOURCE[0]}' called by '${BASH_SOURCE[1]}'"
echo End b.sh
x

Now run the code and review the output:

./a.sh
This is a.sh
This is b.sh
BASH_SOURCE=(b.sh ./a.sh) and we are 'b.sh' called by './a.sh'
End b.sh
End a.sh

As you can see, in a sourced file the caller can be identified with "${BASH_SOURCE[1]}".

Answer 2

It would be possible with the bash shell. source can supply an argument to the file there.

That means you could construct foo.sh as:

#!/bin/bash
echo "Hello from $1"

and bar.sh as

#!/bin/bash
source ~/Codes/tests/foo.sh '~/bar.sh'

finally baz.sh would look like this:

#!/bin/bash
source ~/Codes/tests/foo.sh '~/baz.sh'

If you care about how the script was invoked, then you could also write foo.sh as

#!/bin/bash
echo "Hello from $0"

and bar.sh as

#!/bin/bash
source ~/Codes/tests/foo.sh

This will give Hello from ./bar.sh if you invoke it from the scripts directory. If you invoke it from your home you will get Hello from ~/bar.sh