Zip function for shell scripts

Question

I'm trying to write a shell script that will make several targets into several different paths. I'll pass in a space-separated list of paths and a space-separated list of targets, and the script will make DESTDIR=$path $target for each pair of paths and targets. In Python, my script would look something like this:

for path, target in zip(paths, targets):
    exec_shell_command('make DESTDIR=' + path + ' ' + target)

However, this is my current shell script:

#! /bin/bash

packages=$1
targets=$2
target=

set_target_number () {
    number=$1
    counter=0
    for temp_target in $targets; do
        if [[ $counter -eq $number ]]; then
            target=$temp_target
        fi
        counter=`expr $counter + 1`
    done 
}

package_num=0
for package in $packages; do
    package_fs="debian/tmp/$package"
    set_target_number $package_num
    echo "mkdir -p $package_fs"
    echo "make DESTDIR=$package_fs $target"
    package_num=`expr $package_num + 1`
done

Is there a Unix tool equivalent to Python's zip function or an easier way to retrieve an element from a space-separated list by its index? Thanks.

Answer 1

There's no way to do that in bash. You'll need to create two arrays from the input and then iterate through a counter using the values from each.

Answer 2

Use an array:

#!/bin/bash
packages=($1)
targets=($2)
if (("${#packages[@]}" != "${#targets[@]}"))
then
    echo 'Number of packages and number of targets differ' >&2
    exit 1
fi
for index in "${!packages[@]}"
do
    package="${packages[$index]}"
    target="${targets[$index]}"
    package_fs="debian/tmp/$package"
    mkdir -p "$package_fs"
    make "DESTDIR=$package_fs" "$target"
done