How to use getopts with single character and multicharacter switches

#!/bin/bash

# A small example program for using the new getopt(1) program.
# This program will only work with bash(1)
# An similar program using the tcsh(1) script language can be found
# as parse.tcsh

# Example input and output (from the bash prompt):
# ./parse.bash -a par1 'another arg' --c-long 'wow!*\?' -cmore -b " very long "
# Option a
# Option c, no argument
# Option c, argument `more'
# Option b, argument ` very long '
# Remaining arguments:
# --> `par1'
# --> `another arg'
# --> `wow!*\?'

# Note that we use `"$@"' to let each command-line parameter expand to a 
# separate word. The quotes around `$@' are essential!
# We need TEMP as the `eval set --' would nuke the return value of getopt.
TEMP=`getopt -o ab:c:: --long a-long,b-long:,c-long:: \
     -n 'example.bash' -- "$@"`

if [ $? != 0 ] ; then echo "Terminating..." >&2 ; exit 1 ; fi

# Note the quotes around `$TEMP': they are essential!
eval set -- "$TEMP"

while true ; do
 case "$1" in
  -a|--a-long) echo "Option a" ; shift ;;
  -b|--b-long) echo "Option b, argument \`$2'" ; shift 2 ;;
  -c|--c-long) 
   # c has an optional argument. As we are in quoted mode,
   # an empty parameter will be generated if its optional
   # argument is not found.
   case "$2" in
    "") echo "Option c, no argument"; shift 2 ;;
    *)  echo "Option c, argument \`$2'" ; shift 2 ;;
   esac ;;
  --) shift ; break ;;
  *) echo "Internal error!" ; exit 1 ;;
 esac
done
echo "Remaining arguments:"
for arg do echo '--> '"\`$arg'" ; done

How to replace a string in a string variable in bash without using sed, awk or perl

Bash provides a way to replace a string
${<STRING>//<PATTERN TO FIND IN STRING>/<STRING TO REPLACE>}