How to process in a script Bash arguments with spaces?

Question

I have a Bash script which (in a simplified form) does this:

#!/bin/bash

function ag_search_and_replace {
  ag -l "$1" "${@:3}" | xargs -r perl -i -pe "s/$1/$2/g"
}

locations="$@"

ag_search_and_replace search replace $locations

This works as expected when the argument have no spaces, e.g.:

my_script foo bar

however, when there are spaces, e.g.:

my_script foo "ba r"

the script fails.

Is there a simple way to process arguments with spaces?

chepner · Accepted Answer · 2018-07-02 13:29:19Z

4

"$@" is the way to do it, but you lose the benefits when you unnecessarily assign it to a regular variable.

ag_search_and_replace search replace "$@"

If you must create a new named variable, use an array.

locations=( "$@" )
ag_search_and_replace search replace "${locations[@]}"

answered Jul 2, 2018 at 13:29

chepner

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4 Comments

Marcus Over a year ago

The assignment to regular variable is necessary in the context of the full script :-) This is a simplified version; I've put the assignment for consistency with the script.

chepner Over a year ago

Then the full script is broken.

Marcus Over a year ago

What do you refer, specifically, with broken (I assume it's something else other than what you corrected by the reply)?

chepner Over a year ago

Because location="$@" and location="$*" do exactly the same thing, and neither one lets you retrieve the original arguments from location. If location='a b c', then did you originally have 1) a b c, 2) a b and c, 3) a and b c, or 4) a, b, and c?

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