Question

If inner.sh is

#...
echo first
echo second
echo third

And outer.sh is

var=`./inner.sh`
# only wants to use "first"...  

How can var be split by whitespace?

Answer 1

Try this:

var=($(./inner.sh))

# And then test the array with:

echo ${var[0]}
echo ${var[1]}
echo ${var[2]}

Output:

first
second
third

Explanation:

• You can make an array in bash by doing var=(first second third), for example.
• $(./inner.sh) runs the inner.sh script, which prints out first, second, and third on separate lines. Since we don’t didn’t put double quotes around $(...), they get lumped onto the same line, but separated by spaces, so you end up with what it looks like in the previous bullet point.