Exit a bash switch statement - bash

Question

I've written a menu driven bash script that uses a switch case inside of a while loop to perform the various menu options. Everything works just fine. Now I'm trying to improve the program by performing error testing on the user input, but I cannot seem to make it work...
The problem is I don't know how to properly break out of a switch statement, without breaking out of the while loop (so the user can try again).
# repeat command line indefinitely until user quits
while [ "$done" != "true" ]
do
# display menu options to user
echo "Command Menu" # I cut out the menu options for brevity....
# prompt user to enter command
echo "Please select a letter:"
read option
# switch case for menu commands, accept both upper and lower case
case "$option" in
# sample case statement
a|A) echo "Choose a month"
read monthVal
if [ "$monthVal" -lt 13 ]
then
cal "$monthVal"
else
break # THIS DOES NOT WORK. BREAKS WHILE LOOP, NOT SWITCH!
fi
;;
q|Q) done="true" #ends while loop
;;
*) echo "Invalid option, choose again..."
;;
esac
done
exit 0
The program works fine when the user enters a valid month value, but if they enter a number higher than 13, instead of breaking the switch statement and repeating the loop again, the program breaks both the switch and while loop and stops running.

Answer 1

Hitting ;; will terminate the case statement. Try not doing anything at all:
a|A) echo "Choose a month"
read monthVal
if [ "$monthVal" -lt 13 ]
then
cal "$monthVal"
fi
;;

Answer 2

Move the body of that case into a function, and you can return from the function at will.
do_A_stuff() {
echo "Choose a month"
read monthVal
if [ "$monthVal" -lt 13 ]
then
cal "$monthVal"
else
return
fi
further tests ...
}
Then
case $whatever in
a|A) do_A_stuff ;;

Answer 3

I think what you mean to do with break is "quit this case statement and restart the while loop". However, case ... esac isn't a control-flow statement (although it may smell like one), and doesn't pay attention to break.
Try changing break into continue, which will send control back to the start of the while loop.

Answer 4

This should do the trick: Wrap the code in a one-trip for-loop:
#! /bin/bash
case foo in
bar)
echo "Should never get here."
;;
foo)
for just in this_once ; do
echo "Top half only."
if ! test foo = bar; then break; fi
echo "Bottom half -- should never get here."
done
;;
*)
echo "Should never get here either."
;;
esac

Answer 5

In your example, there is no point in breaking, you can omit the else break statement altogether.
The problem occurs when there is code that runs after the point where you would have broken. You'd want to write something like that
case $v in
a) if [ $x ]; then bla; else break; fi;
some more stuff ;;
b) blablabla ;;
What I usually do (because creating a function is such a hassle with copy pasting, and mostly it breaks the flow of the program when you read it to have a function somewhere else) is to use a break variable (which you can call brake for fun when you have a lame sense of humor like me) and enclose "some more stuff" in an if statement
case $v in
a) if [ $x ]; then bla; else brake="whatever that's not an empty string"; fi;
if [ -z "$brake" ];then some more stuff; brake=""; fi;;
#don't forget to clear brake if you may come back here later.
b) blablabla ;;
esac