How to iterate a command with two different variables?

Question

Using kernel 2.6.x

GNU bash, version 4.3.42(1)-release (arm-openwrt-linux-gnu)

In a bash script, how would you write a for loop that produces the following result with the variables below ?

command option option abc option 10
command option option def option 20

VAR1="abc def"

VAR2="10 20"

I can iterate $VAR1, but I don't know how to iterate $VAR2.

for i in $VAR1; do
command option option "$VAR1" option "$VAR2"
done

Should the command line be split into two strings, iterated separately and then concatenated ?

How about using a count variable with an array ? For example ...

count=1
echo ${VAR1["$count"]}

Could a count variable with a while loop work ?

Answer 1

How about this:

arr1=(abc def)
arr2=(10 20)

And then:

for (( i=0; i<${#arr1[@]}; i++ )); do echo "${arr1[i]} ${arr2[i]}"; done
abc 10
def 20

In your case, the echo statement will then look like:

for (( i=0; i<${#arr1[@]}; i++ )); do 
    echo "command option option ${arr1[i]} option ${arr2[i]}"
done
command option option abc option 10
command option option def option 20

Look here for examples on a C style for loop in bash.

Answer 2

It's easier with zsh:

$ a=(a b c) b=(1 2 3)
$ for i j (${a:^b}) echo "i: $i, j: $j"
i: a, j: 1
i: b, j: 2
i: c, j: 3

${a:^b} is the array-zipping operator (expands to a, 1, b, 2, c, 3 above).

If you wanted to loop over words resulting of the split+glob operator applied to $VAR1 and $VAR2 in a POSIX-like shell (like the busybox sh you seem to be using according to the comments to @maulinglawns's answer), you could use the positional parameters.

VAR1='a b c'
VAR2='1 2 3'
set -- $VAR1
for j in $VAR2; do
  echo "i: $1, j: $j"
  shift
done