bash if loop for checking multiple parameters

Shell Programming and Scripting
1

Hello,

I've got next problem:
I want to examine at the beginning of a script in an if loop that:

  1. Is there 4 parameters given
  2. If first state is true then: is there switches -e and -d?
  3. At the end, how can i indentify them as variebles regardlees to its order.

I was thinking like this if [ "$#" == 4 && "$1" == -e && "$3" == -d ] ; then ...

Can anyone help me?
If loop could be here the right choice, at all?

Regards

2

You can use getopts here.

e.g.

[ $# -ne 4 ] && <some usage function> && exit

while getopts e:d: OPTION
do
    case $OPTION in
        e)  somevar=$OPTARG ;;
        d)  somevar1=$OPTARG ;;
    esac
done
3

There is no such thing as an if loop; if chooses between one or more sets of commands to execute depening on one or more conditions.

if [ $# -eq 4 ]
then
   : do whatever
else
  printf "4 parameters required\n" >&2
  exit 1 ## or whatever
fi

They are called options, not switches.

Regard what as variables?

As jaduks pointed out, the usual method is to use getopts in a while loop.

However, it is easy enough to roll your own:

while [ $# -gt 0 ]
do
  case $1 in
    -e) var_e=$2
        shift 2
        ;;
    -d) var_d=$2
        shift 2
        ;;
     *) shift ;; ## ignore other arguments
  esac
done