What is the most concise way to source a file (only if it exists) in Bash?

Question

In Bash scripting, is there a single statement alternative for this?

if [ -f /path/to/some/file ]; then
    source /path/to/some/file
fi

The most important thing is that the filename is there only once, without making it a variable (which adds even more lines).

For example, in PHP you could do it like this

@include("/path/to/some/file"); // @ makes it ignore errors

Solution 1

Is defining your own version of @include an option?

include () {
    [[ -f "$1" ]] && source "$1"
}

include FILE

Solution 2

If you're concerned about a one-liner without repeating the filename, perhaps:

FILE=/path/to/some/file && test -f $FILE && source $FILE

Solution 3

If you are concerned about warning (and lack of existence of sourced file isn't critical for your script) just get rid of the warning:

source FILE 2> /dev/null

Solution 4

If you want to always get a clean exit code, and continue no matter what, then you can do:

source ~/.bashrc || true && echo 'is always executed!'

And if you also want to get rid of the error message then:

source ~/.bashrc 2> /dev/null || true && echo 'is always executed!'

Solution 5

You could try

test -f $FILE && source $FILE

If test returns false, the second part of && is not evaluated

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