getopt erroneously caches arguments - bash

Question

I've created a script in my bash_aliases to make SSH'ing onto servers easier. However, I'm getting some odd behavior that I don't understand. The below script works as you'd expect, except for when it's re-used.
If I use it like this for this first time in a shell, it works exactly as expected:
$>sdev -s myservername
ssh -i ~/.ssh/id_rsa currentuser#myservername.devdomain.com
However, if I run that a second time, without specifying -s|--server, it will use the server name from the last time I ran this, having seemingly cached it:
$>sdev
ssh -i ~/.ssh/id_rsa currentuser#myservername.devdomain.com
It should have exited with an error and output this message: /bin/bash: A server name (-s|--server) is required.
This happens with any of the arguments; that is, if I specify an argument, and then the next time I don't, this method will use the argument from the last time it was supplied.
Obviously, this is not the behavior I want. What's responsible in my script for doing that, and how do I fix it?
#!/bin/bash
sdev() {
getopt --test > /dev/null
if [[ $? -ne 4 ]]; then
echo "`getopt --test` failed in this environment"
exit 1
fi
OPTIONS=u:,k:,p,s:
LONGOPTIONS=user:,key:,prod,server:
# -temporarily store output to be able to check for errors
# -e.g. use “--options” parameter by name to activate quoting/enhanced mode
# -pass arguments only via -- "$#" to separate them correctly
PARSED=$(getopt --options=$OPTIONS --longoptions=$LONGOPTIONS --name "$0" -- "$#")
if [[ $? -ne 0 ]]; then
# e.g. $? == 1
# then getopt has complained about wrong arguments to stdout
exit 2
fi
# read getopt’s output this way to handle the quoting right:
eval set -- "$PARSED"
domain=devdomain
user="$(whoami)"
key=id_rsa
# now enjoy the options in order and nicely split until we see --
while true; do
case "$1" in
-u|--user)
user="$2"
shift 2
;;
-k|--key)
key="$2".pem
shift 2
;;
-p|--prod)
domain=proddomain
shift
;;
-s|--server)
server="$2"
shift 2
;;
--)
shift
break
;;
*)
echo "Programming error"
exit 3
;;
esac
done
if [ -z "$server" ]; then
echo "$0: A server name (-s|--server) is required."
kill -INT $$
fi
echo "ssh -i ~/.ssh/$key.pem $user#$server.$domain.com"
ssh -i ~/.ssh/$key $user#$server.$domain.com
}

Answer 1

server is a global shell variable, so it's shared between runs of the function (as long as they're run in the same shell). That is, when you run sdev -s myservername, it sets the variable server to "myservername". Later, when you run just sdev, it checks to see if $server is empty, finds it's not, and goes ahead and uses it.
Solution: use local variables! Actually, it'd be best to declare all of the variables you use in the function as local; that way, you don't run the risk of interfering with something else that's trying to use the same variable name. I'd also recommend avoiding all-caps variable names (like OPTIONS, LONGOPTIONS, and PARSED) -- there are a bunch of all-caps variables that have special meanings to the shell and/or other programs, and if you use one of those by mistake it can cause weird problems.
Anyway, here's the simple solution: add this near the beginning of the script:
local server=""