LinuxQuestions.org - Using getopts in Shell properly??

I have a program which is basically written in sh (has to be) and it is trying to read inputs from the command line and if no input is given then it takes the default value.

Here is the code

Code:

#!/bin/sh



if [ -f /.default ]

then

. /.default

fi



#Set Default Values from the file

DN=$DEFAULTDN

PD=$DEFAULTPD

RPD=$DEFAULTRPD

HOSTNAME=`hostname`



while getopts D:w:r:h:p:i:t: name

do

  case $name in

    t)  TYPE=$OPTARG;;

    D)  DN=$OPTARG;;

    w)  PD=$OPTARG;;

    r)  RPD=$OPTARG;;

    h)  HOSTNAME=$OPTARG;;

    p)  if [ $TYPE = option1 ]

        then

        NUM=747

        elif [ $TYPE = option2 ]

        then

        NUM=777

        fi

        ;;

    i)  if [ $TYPE = option1 ] ; then

                IPATH='/net/install/$HOSTNAME'

        elif [ $TYPE = option2 ] ; then

                IPATH='/net/install2/$HOSTNAME-opt2'

        fi

        ;;



    ?)  usage

        exit 1

        ;;

  esac

done



echo $TYPE

echo $DN

echo $PDN

echo $HOSTNAME

echo $NUM

echo $IPATH

Basically if the file is called test.sh I want the user to run it with mininum commands

Code:

./test -t (option1|option2)

Once he specifies that, and if all the other are not specified (i.e. D w r h etc) then it will take the default.. If even -D -w -r etc are mentioned then it overrides the default value

This script is not working. It is sourcing correctly from the file and assigning the default value and echo out is correct. But I think because I am not mentioning -D -w -r etc , the getopts does not work

Can anyone suggest how to get this working??

Ta