How do i use options in bash shell scripts?

vaquitasos

The options to various commands are the -a, -p, -g type syntax you see after a command

For example -u is an option.

$ who -u

Shell scripts can also have these options.

Some info on the commands used in the script:

       shift [n]
              The positional parameters from n+1 ... are renamed  to  $1  ....
              Parameters  represented  by  the  numbers  $# down to $#-n+1 are
              unset.  n must be a non-negative number less than  or  equal  to
              $#.   If  n is 0, no parameters are changed.  If n is not given,
              it is assumed to be 1.  If n is greater than $#, the  positional
              parameters  are  not changed.  The return status is greater than
              zero if n is greater than $# or less than zero; otherwise 0.

       break [n]
              Exit from within a for, while, until, or select loop.  If  n  is
              specified, break n levels.  n must be ≥ 1.  If n is greater than
              the number of enclosing loops, all enclosing loops  are  exited.
              The  return  value is 0 unless n is not greater than or equal to
              1.

       --        A -- signals the end of options and disables  further  option
                 processing.   Any arguments after the -- are treated as file‐
                 names and arguments.  An argument of - is equivalent to --.

Here's a script that passes three options, -a, -b, and -c, option b has a parameter.

#!/bin/bash
while [ -n "$1" ]
do
  case "$1" in
  -a) echo "The user type option -a";;
  -b) parameter="$2"
  echo "The user typed option -b with a parameter of $parameter"
  shift ;;
  -c) echo "The user types option -c";;
  --) shift
  break ;;
  *) echo "That option $1 does not exist";;
  esac
  shift
done

The shift command visually.