Bash optarg fails to spot missing arguments - bash

Question

I am inexperienced with bash shell scripting, and have run into a problem with bash optarg
Here's a small script to reproduce the problem:
#!/bin/sh
while getopts ":a:b:" opt; do
case ${opt} in
a ) echo "a=$OPTARG"
;;
b ) echo "b=$OPTARG"
;;
\? ) echo "Invalid option: $OPTARG" 1>&2
;;
: ) echo "Invalid option: $OPTARG requires an argument" 1>&2
esac
done
When I try this:
./args.sh -a av -b bv
I get the expected result:
a=av
b=bv
But when I omit the argument for -a:
/args.sh -a -b bv
I get this unfortunate result:
a=-b
When I would expect an error to show that the value of -a is missing.
It seems to have taken the -b argument as the value for -a.
Have I done something wrong & how can I achieve the expected behaviour?

Answer 1

The only positive advice is how do you treat But when I omit the argument for '-a', you cannot just skip to the next subsequent option. By convention getopts a: means you are expecting to an provide an arg value for the flag defined.
So even for the omitting case, you need to define an empty string which means the value for the arg is not defined i.e.
-a '' -b bv
Or if you don't expect the -a to get any arg values, better change the option string to not receive any as :ab:.
Any other ways of working around by checking if the OPTARG for -a is does not contain - or other hacks are not advised as it does not comply with the getopts() work flow.

Answer 2

getopts doesn't support such detection. So there's no way to do that with getopts.
You can probably write a loop around the arguments instead. something like:
#!/bin/sh
check_option()
{
case $1 in
-*)
return 1
;;
esac
return 0
}
for opt in $#; do
case ${opt} in
-a) shift
if check_option $1; then
echo "arg for -a: $1"
shift
else
echo "Invalid option -a"
fi
;;
-b) shift
if check_option $1; then
echo "arg for -b: $1"
shift
else
echo "Invalid option -b"
fi
;;
esac
done