Dealing with multiple flags via getopt - bash

Question

I am using a script which should be invoked using only a single parameter to each flag as follows:
./testit.sh -n 123 -t tvar -b bvar -s svar my_program.exe flags to my program
where all flags are just flags to the script, and the script will launch my_program.exe with the flags flags to my program. The simple method being used to do this is getopts as follows:
#!/bin/bash
# contents of "test_it.sh"
echo "\$* = "$*
getopt_out=`getopt t:T:b:B:s:S:n:N $*`
echo "getopt_out = "$getopt_out
echo "\$* = "$*
set -- `getopt t:T:b:B:s:S:n:N $*`
echo "\$* = "$*
while [ $1 != -- ]; do
shift
done
shift
echo "**********************************************************************************"
echo "*** The program I want to run is: "$*
echo "**********************************************************************************"
where the (correct) output is:
**********************************************************************************
*** The program I want to run is: my_program.exe flags to my program
**********************************************************************************
I, however, need to send multiple numerical options to -n as follows:
./testit.sh -n 123 456 789 -t tvar -b bvar -s svar my_program.exe flags to my program
which gives incorrect output:
**********************************************************************************
*** The program I want to run is: 456 789 my_program.exe flags to my program
**********************************************************************************
How can I, from within 'testit.sh', get rid of those extra numbers? I am able to deal with them (log their values) at the beginning of the script so that they are no longer needed. Since the testit.sh script relies on shift, is there a way for me to completely disregard (throw out) the numeric values without messing up the flow of the commands, so that shift can still be used?
EDIT: Upon further investigation of my original script, I noticed that the output of getopts is different than the one posted in my minimal example. I have updated the minimal example along with the proposed workaround in an answer, though I would appreciated other (probably more canonical and/or correct) methods for dealing with this).

Answer 1

It turns out the original script which was causing me problems actually used POSIXLY_CORRECT=1 which slightly changes the order of -- in the output. I have posted the new code (including POSIXLY_CORRECT=1), along with my workaround for anyone who is interested:
#!/bin/sh
echo "\$* = "$*
POSIXLY_CORRECT=1
export POSIXLY_CORRECT
getopt_out=`getopt t:T:b:B:s:S:n:N: $*`
echo "getopt_out = "$getopt_out
set -- `getopt t:T:b:B:s:S:n:N: $*`
echo "\$* = "$*
double_dash_too_soon=false
while [ $1 != -- ]; do
echo "\$1 = "$1
case $1 in
-t)
# process this flag
shift
;;
-b)
# process this flag
shift
;;
-s)
# process this flag
shift
;;
-n)
This_num=$2
echo "This_num = "$This_num
shift
while [[ $2 != "-"[a-z,A-Z] ]] ; do
echo "Dealing with number!"
if [[ $1 == "--" ]]; then
echo "WARNING: '--' encountered during numeric reads!!" >&2
double_dash_too_soon=true
fi
shift
done
;;
esac
shift
if [ "$double_dash_too_soon" == true ]; then
double_dash_too_soon=false
echo "(1) \$* -> "$*
set -- `getopt t:T:b:B:s:S:n:N $*`
echo "(2) \$* -> "$*
fi
done
shift
echo "**********************************************************************************"
echo "*** The program I want to run is: "$*
echo "**********************************************************************************"