8

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?

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1
  • You can first replace your source $(cat) with read file; source $file. That's the usual way to read a word from standard input. Commented Aug 14, 2014 at 16:12

1 Answer 1

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9

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
}
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2
  • 1
    Boom. That's the answer. Extra kudos for predicting the multi-file use case which I was actually trying to support. For posterity's sake, you might want to remove from the source code snippets the comment about "getting from stdin" since we're no longer doing that. Commented Aug 14, 2014 at 18:41
  • @cuonglm so is totally valid do a source within a method? Commented Feb 17, 2022 at 13:45

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