How to call a script with a var interpreted as multiple arguments

Question

I have a var which is a string that contains several words like this:

my_var="foo bar baz"

I'd like to call another script and pass in my_var. I want this second script to see three arguments foo bar baz instead of a single argument "foo bar baz"

How is this done? Here's what currently have:

text_output=$( /bin/bash /the/second/script.sh ${my_var} )

This second script is expecting any number of arguments. And currently it says it is getting one argument. Could the syntax ${} be the issue?

If you aren't quoting ${my_var}, something in your script is miscounting the number of arguments it is receiving. — chepner
– chepner, Commented May 1, 2015 at 18:43
Either that, or IFS got changed and that's confusing things. — Gordon Davisson
– Gordon Davisson, Commented May 2, 2015 at 5:30

chepner · Accepted Answer · 2015-05-01 18:41:59Z

2

Don't use a plain parameter to store multiple arguments; this is what arrays are for.

declare -a my_arr    # optional
my_arr=( foo bar baz )
text_output=$( /bin/bash /the/second/script.sh "${my_arr[@]}" )

answered May 1, 2015 at 18:41

chepner

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anubhava · Accepted Answer · 2015-05-01 18:41:52Z

0

You can pass it unquoted so that shell will be able to expand it. Examples:

my_var="foo bar baz"

fn() { echo $#; }

fn "$my_var"
1

fn $my_var
3

answered May 1, 2015 at 18:41

anubhava

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