How to get shellscript filename without $0?

Question

I tried to use python practice if __name__ == "__main__": on shellscript.

Sample scripts are the following:

a.sh:

#!/bin/bash

filename="a.sh"

function main() {
  echo "start from $0"
  echo "a.sh is called"
  source b.sh
  bfunc
}

[[ "$0" == "${filename}" ]] && main

b.sh:

#!/bin/bash

filename="b.sh"

function main() {
  echo "start from $0"
  echo "b.sh is called"
}

function bfunc() {
  echo "hello bfunc"
}

[[ "$0" == "${filename}" ]] && main

You can call it with bash a.sh.

If you call bash a.sh, you'll get like the following:

start from a.sh
a.sh is called
hello bfunc

Here is my question. How can I get file name itself without using $0? I don't want to write file name directly, i.e. I want to pass the file name value to ${filename}.

See the link if you don't know that is the above python practice: What does if __name__ == "__main__": do?

How can I check wheather b.sh is started from command line or was executed by including from a.sh?

Answer 1

You may use the variable $BASH_SOURCE to get the name of the current script file.

if [[ "$0" == "$BASH_SOURCE" ]]
then
    : "Execute only if started from current script"
else
    : "Execute when included in another script"
fi