Extract parameters before last parameter in "$@"

Question

I'm trying to create a Bash script that will extract the last parameter given from the command line into a variable to be used elsewhere. Here's the script I'm working on:

#!/bin/bash # compact - archive and compact file/folder(s)  eval LAST=\$$#  FILES="$@" NAME=$LAST  # Usage - display usage if no parameters are given if [[ -z $NAME ]]; then   echo "compact <file> <folder>... <compressed-name>.tar.gz"   exit fi  # Check if an archive name has been given if [[ -f $NAME ]]; then   echo "File exists or you forgot to enter a filename.  Exiting."   exit fi  tar -czvpf "$NAME".tar.gz $FILES 

Since the first parameters could be of any number, I have to find a way to extract the last parameter, (e.g. compact file.a file.b file.d files-a-b-d.tar.gz). As it is now the archive name will be included in the files to compact. Is there a way to do this?

Answer 1

To remove the last item from the array you could use something like this:

#!/bin/bash  length=$(($#-1)) array=${@:1:$length} echo $array 

Even shorter way:

array=${@:1:$#-1} 

But arays are a Bashism, try avoid using them :(.

Answer 2

Portable and compact solutions

This is how I do in my scripts

last=${@:$#} # last parameter  other=${*%${!#}} # all parameters except the last 

EDIT
According to some comments (see below), this solution is more portable than others.
Please read Michael Dimmitt's commentary for an explanation of how it works.