Having bash script as follows
#! /bin/bash
usage()
{
echo -e "need help!"
}
while getopts ":a:b:h" OPTION
do
case $OPTION in
a)
printf "a option with value %sn" $OPTARG
;;
b)
printf "b option with value %sn" $OPTARG
;;
h)
usage
;;
?)
echo -e "No option selected"
;;
esac
done
exit 0
Above script run fine with different option but i wanted to extend it to not allow to pass multiple option at same time like as following argument
$ ./test.bash -a 1 -b 2 a option with value 1 b option with value 2
should be not valid means some way it give me error like wrong syntax i achieved it by as follows but it seems to long it is as follow
#! /bin/bash
usage()
{
echo -e "need help!"
}
let "a_count=0"
let "b_count=0"
MY_ARG=""
while getopts ":a:b:h" OPTION
do
case $OPTION in
a)
let a_count=1
MY_ARG=$OPTARG
;;
b)
let b_count=1
MY_ARG=$OPTARG
;;
h)
usage
;;
?)
echo -e "No option selected"
;;
esac
done
[[ $a_count -eq 1 ]] && [[ $b_count -eq 1 ]] && echo "wrong command sytax" && exit 0
[[ $a_count -eq 1 ]] && printf "a option with value %sn" $MY_ARG
[[ $b_count -eq 1 ]] && printf "b option with value %sn" $MY_ARG
exit 0
run like
$ ./test.bash -a 1 -b 2 wrong command sytax
But i want to finish validation in while..loop of getopts. Also this validation not works for following command
./test.bash -a -b a option with value -b
any one have batter idea how to use getopts for this type validation?
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Answer
you’ve almost got it. The : after the a and after the b say that they take an argument, so your example with -a -b is actually valid, saying “There is option a with value -b”.
If you really just want “-a or -b and then an argument”, you probably don’t need getopts at all, but should do:
[ "$1" == "-a" ] && printf "a option with value %sn" $2 [ "$1" == "-b" ] && printf "b option with value %sn" $2