How to use getopts in any random order with long optional or non-optional argument first in bash?

Question

I have a script like

./sample.sh --add employeename -f firstname -a age -d detail
./sample.sh --add employeename  -a age -d detail -f firstname
./sample.sh --add employeename -d detail -f firstname
./sample.sh --del employeename
./sample.sh --update employeename -f firstname [age,detail are optional]

here,how could i use getopts for --add, --del, --update and i don't want to use getopt. Good suggestion will be appreciated.

Answer 1

Those are really commands, not options, so should be regular positional arguments:

./sample.sh add employeename -f firstname -a age -d detail
./sample.sh add employeename  -a age -d detail -f firstname
./sample.sh add employeename -d detail -f firstname
./sample.sh del employeename
./sample.sh update employeename -f firstname [age,detail are optional]

which should come first; then the remaining options can be parsed using a command-specific set of options.

cmd=$1
shift
case $cmd in
  add) do_add "$@" ;;
  del) do_del "$@" ;;
  update) do_update "$@" ;; 
  *) echo "Unrecognized command: $cmd"
     exit 1
     ;;
esac

do_add () {
    name=$1
    shift
    while getopts "f:a:d:" opt "$@"; do
      ...
    done
    ...
}

do_del () {
    ...
}

# etc