How can I send multiple options to one variable

Question

I am running a loop for an array of variables that generates three more variables.

i.e

for foo in ${bar[@]};
    do bar1=$(echo $foo | awk '{print$1}');
       bar2=$(echo $foo | awk '{print$2}');
       bar3=$(echo $foo | awk '{print$3}');
done

However when the loop is finished I need the three newly created variables to become arrays themselves that I can run for loops on as well.

i.e.

for foo in ${bar1[@]}
    do echo $foo
done

^^^^this should show multiple lines.

can anyone provide anyway to make this possible?

Answer 1

Use

      bar1+=($(echo $foo | awk '{print$1}'))

i.e., variable+=( value ) to add a value to an array.  Of course the code to use the bar1 array has to look something like

for foo in "${bar1[@]}"
do
    echo "$foo"
done

Of course you should always quote shell variables (e.g., "$foo" and "${bar1[@]}") unless you have a good reason not to, and you’re sure you know what you’re doing. 

Answer 2

No need for three pipelines here; just use read to extract the first three fields from the string.

for foo in "${bar[@]}"; do
    read bar1 bar2 bar3 the_rest <<< "$foo"
done

the_rest is only necessary if there's a chance that foo could be something like "a b c d" and you don't want bar3 to be set to c d.