Bash script arguments

Question

Im trying to pass arguments to a script i wrote, but cant get it right.

What i want is one mandatory argument without a flag, and two optional arguments with flags, so it can be called like this:

./myscript mandatory_arg -b opt_arg -a opt_arg

or

./myscript mandatory_arg -a opt_arg
./myscript mandatory_arg -b opt_arg

I looked into getopts and got this:

while getopts b:a: option
do
    case "${option}"
    in
        b) MERGE_BRANCH=${OPTARG};;
        a) ACTION=${OPTARG};;
    esac
done

if "$1" = ""; then
    exit
fi

echo "$1"
echo "$MERGE_BRANCH"
echo "$ACTION"

But it does not work at all.

Patrick Trentin · Accepted Answer · 2017-02-03 09:57:23Z

3

Assuming that your mandatory argument appears last, then you should try the following code: [comments inline]

OPTIND=1
while getopts "b:a:" option
do
    case "${option}"
    in
        b) MERGE_BRANCH=${OPTARG};;
        a) ACTION=${OPTARG};;
    esac
done

# reset positional arguments to include only those that have not
# been parsed by getopts

shift $((OPTIND-1))
[ "$1" = "--" ] && shift

# test: there is at least one more argument left

(( 1 <= ${#} )) || { echo "missing mandatory argument" 2>&1 ; exit 1; };

echo "$1"
echo "$MERGE_BRANCH"
echo "$ACTION"

The result:

~$ ./test.sh -b B -a A test
test
B
A
~$ ./tes.sh -b B -a A
missing mandatory argument

If you really want the mandatory argument to appear first, then you can do the following thing:

MANDATORY="${1}"
[[ "${MANDATORY}" =~ -.* ]] && { echo "missing or invalid mandatory argument" 2>&1; exit 1; };

shift # or, instead of using `shift`, you can set OPTIND=2 in the next line   
OPTIND=1
while getopts "b:a:" option
do
    case "${option}"
    in
        b) MERGE_BRANCH=${OPTARG};;
        a) ACTION=${OPTARG};;
    esac
done

# reset positional arguments to include only those that have not
# been parsed by getopts

shift $((OPTIND-1))
[ "$1" = "--" ] && shift

echo "$MANDATORY"
echo "$MERGE_BRANCH"
echo "$ACTION"

The result is the following:

~$ ./test.sh test -b B -a A
test
B
A
~$ ./tes.sh -b B -a A
missing or invalid mandatory argument

edited Feb 3, 2017 at 9:57

answered Feb 3, 2017 at 9:04

Patrick Trentin

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5 Comments

A.Jac Over a year ago

When trying to run this it prints out the mandatory argument and then "" for the optionals, if i run without the mandatory it replies with "missing mandatory argument"

A.Jac Over a year ago

What is the reason for not recommending having the mandatory argument first? and i think something is missing from the last edit

Patrick Trentin Over a year ago

@A.Jac it's just a matter of convention and personal taste

Patrick Trentin Over a year ago

@A.Jac Assume that your mandatory argument is the name of a file, and that your file name is -test, with the former code you can write ./test..sh -b B -a A -- -test and have your program work correctly, whereas with the last code you won't be able to parse it: -- is used to make getopt stop looking for options, but A) -b B -a A won't be recognized as options B) -- will fail the test [[ ${MANDATORY} =~ -.* ]]

Patrick Trentin Over a year ago

Anyway I removed my personal opinions from the answer, because clearly there are workarounds for the issues I see with this approach, it is only a matter of implementing them with some more extra code.

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