[SOLVED] How to split up function from arguments in bash?

Issue

I am writing a bash script named safeDel.sh with base functionalities including:

• file [file1, file2, file3...]
• -l
• -t
• -d
• -m
• -k
• -r arg

For the single letter arguments I am using the built in function getops which works fine. The issue I'm having now is with the 'file' argument. The 'file' argument should take a list of files to be moved to a directory like this:

$ ./safeDel.sh file file1.txt file2.txt file3.txt

The following is a snippet of the start of my program :

#! /bin/bash

files=("$@")
arg="$1"
echo "arguments: $arg $files" 

The echo statement shows the following:

$ arguments : file file

How can I split up the file argument from the files that have to be moved to the directory?

Solution

Assuming that the options processed by getopts have been shifted off the command line arguments list, and that a check has been done to ensure that at least two arguments remain, this code should do what is needed:

arg=$1
files=( "${@:2}" )

echo "arguments: $arg ${files[*]}"

files=( "${@:2}" ) puts all the command line arguments after the first into an array called files. See Handling positional parameters [Bash Hackers Wiki] for more information.

${files[*]} expands to the list of files in the files array inside the argument to echo. To safely expand the list in files for looping, or to pass to a command, use "${files[@]}". See Arrays [Bash Hackers Wiki].

Answered By - pjh

Answer Checked By - Clifford M. (WPSolving Volunteer)