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?

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3 Answers

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.
5

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)
13

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
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