Calling `source` from bash function

Question

I'm trying to source a file whose name is passed from stdin. My plan is to create a function like this:

mySource() {
    # get stdin and pass it as an argument to `source`
    source $(cat)
}

to be called like this: $ echo "file1.sh" | mySource wherein file1.sh is:

FILE=success
export FILE

Assuming $FILE is initialized to hello world, when I run $ echo "file1.sh" | mySource, I expect $ echo $FILE to print success; however, instead it prints hello world.

Is there some way to source a file from a function?

Answer 1

You can change your mySource function to:

mySource() {
  source "$1"
}

Then calling it with:

$ mySource file.sh
$ printf '%s\n' "$FILE"
success

You can also make mySource handles multiple files:

mySource() {
  for f do
    source "$f"
  done
}