Bash shell script to extract archive - arrays

Question

I am trying to convert the following script which I use to create archives into one which extracts them.
[[ $# -lt 2 ]] && exit 1
name=$1; shift
files=("$#")
#exclude all files/directories that are not readable
for index in "${!files[#]}"; do
[[ -r ${files[index]} ]] || unset "files[index]"
done
[[ ${#files[#]} -eq 0 ]] && exit 1
if tar -czvf "${name:-def_$$}.tar.gz" "${files[#]}"; then
echo "Ok"
else
echo "Error"
exit 1
fi
So far I have this:
[[ $# -lt 1 ]] && exit 1
files=("$#")
#remove files and directories which are not readable
for index in "${!files[#]}"; do
[[ -r ${files[index]} ]] || unset "files[index]"
done
[[ ${#files[#]} -eq 0 ]] && exit 1
if tar -xzvf "${files[#]}".tar.gz; then
echo "OK"
else
echo "Error"
exit 1
fi
I dont know whether I needed to keep the shift as for this script I do not need to discard any arguments. I want to be able to take them all and unzip each one. Also I see there is a -C switch which allows the user to choose where the unzipped files go. How would I go about also adding this as an option for the user because they may or may not want to change the directory where the files get unzipped to.

Answer 1

You unfortunately can't just do tar -xzvf one.tar.gz two.tar.gz. Straightforward approach is to use a good old for loop:
for file in "${files[#]}"; do
tar -xzvf "$file"
done
Or you can use this:
cat "${files[#]}" | tar -xzvf - -i
You can have the first argument to be the specified directory for the -C option:
[[ $# -lt 2 ]] && exit 1
target=$1; shift
files=("$#")
#remove files and directories which are not readable
for index in "${!files[#]}"; do
[[ -r ${files[index]} ]] || unset "files[index]"
done
[[ ${#files[#]} -eq 0 ]] && exit 1
mkdir -p -- "$target" || exit 1
for file in "${files[#]}"; do
tar -xzvf "$file" -C "$target"
done
./script /some/path one.tar.gz two.tar.gz
List of files for tar can be also constructed like this:
target=$1; shift
for file; do
[[ -r $file ]] && files+=("$file")
done