Avoid options to be taken as argument automatically - bash

Question

I'd like to do some error checking for a bash script I am writing. In particular, I wish to ensure the following option not to be considered as the argument of an option (intentionally) left empty.
Let's say the following snippet
while getopts “hhelpc1:2:” OPTION
do
case "$OPTION" in
h|help)
usage
exit 1
;;
1)
var1=${OPTARG}
;;
2)
var2=${OPTARG}
;;
c)
test1
;;
esac
done
Assuming my script is called test.sh
By doing something like
./test.sh -1 -2 dddd -c
In the above circumstance test1 output an error message that -2 option is empty. On the opposite, I'd like to raise a warning for -1 being empty, whereas at present -2 will be taken as the argument for -1.
Any help?
Thanks
Andrea

Answer 1

getopts:
only handles short option names, so you cannot put "help" in your option string -- that means you're looking for "-h", "-e", "-l", "-p"
cannot look for missing arguments the way you're hoping. You'll have to examine $OPTARG to check if it looks like one of your options.
Add a leading : to the opt string to handle getopts errors yourself.
Here's a reworking of your code and I'm sure there are plenty of cases I'm not catching
#!/bin/bash
usage () { echo usage ...; }
test1 () { echo test1; }
shopt -s extglob
while getopts ":hc1:2:" opt
do
case $opt in
h)
usage
exit 1
;;
1)
case $OPTARG in
-[hc2]*)
echo "error: required argument missing for -1"
usage
exit 1
;;
*) var1=$OPTARG
;;
esac
;;
2)
case $OPTARG in
-[hc1]*)
echo "error: required argument missing for -2"
usage
exit 1
;;
*) var2=$OPTARG
;;
esac
;;
c)
test1
;;
:)
echo "error: required argument missing for -$OPTARG"
usage
exit 1
;;
\?)
# unknown argument, handle accordingly
;;
esac
done
shift $((OPTIND - 1))
echo "var1=$var1"
echo "var2=$var2"
echo "rest=$*"
Update 2014-01-23
Here's one technique:
do_test1=false
while getopts ...
case $opt in
...
c) do_test1=true ;;
...
esac
done
shift $((OPTIND - 1))
# execute function "test1" if "-c" was given:
$do_test1 && test1