Capture multiline output as array in bash

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?

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.

Don't forget the builtin mapfile. It's definitely the most efficient in your case: If you want to slurp the whole file in an array, the fields of which will be the lines output by ./inner.sh, do

mapfile -t array < <(./inner.sh)

Then you'll have the first row in ${array[0]}, etc...

For more info on this builtin and all the possible options:

help mapfile

If you just need the first line in a variable called firstline, do

read -r firstline < <(./inner.sh)

These are definitely the most efficient ways!

This small benchmark will prove it:

$ time mapfile -t array < <(for i in {0..100000}; do echo "fdjsakfjds$i"; done)

real	0m1.514s
user	0m1.492s
sys	0m0.560s
$ time array=( $(for i in {0..100000}; do echo "fdjsakfjds$i"; done) )

real	0m2.045s
user	0m1.844s
sys	0m0.308s

If you only want the first word (with space delimiter) of the first line output by ./inner.sh, the most efficient way is

read firstword _ < <(./inner.sh)

You're not splitting on whitespaces but rather on newlines. Give this a whirl:

IFS=$'
'

var=$(./echo.sh)
echo $var | awk '{ print $1 }'
unset IFS