How to combine getopts and positional parameters in bash?

user_1473423 0
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I want to use both getopts and positional parameters, but if I pass in a positional parameter to the program the getopts get lost.

directory=$1

while getopts l: flag; do
  case "$flag" in
    l) level=$OPTARG;;
  esac
done

if [ -n "$level" ]; then
  echo "Level exist!"
else
  echo "Level doesn't exist!"
fi

So when I run the program like this:

sh myprogram.sh ~/documents -l 2

I expect:

Level exist!

And instead it returns:

Level doesn't exist!

The thing is, if I run the program without the positional parameter (~/documents) like this:

sh myprogram.sh -l 2

I get the correct output:

Level exist!

Why is that? How can I use both positional parameters and getopts in bash?

Thanks!

getopts parameters shell bash
1 Answers

Most tools are written in the form: tool [options] arg ...

So you would do this:

# first, parse the options:
while getopts l: flag; do
  case "$flag" in
    l) level=$OPTARG;;
    \?) exit 42;;
  esac
done

# and shift them away
shift $((OPTIND - 1))

# validation
if [ -n "$level" ]; then
  echo "Level exist!"
else
  echo "Level doesn't exist!"
fi

# THEN, access the positional params
echo "there are $# positional params remaining"
for ((i=1; i<=$#; i++)); do
  printf "%d\t%s\n" $i "${!i}"
done

Use the \? to abort the script if the user provides an unknown option or fails to provide a required argument

And invoke it like:

$ bash test.sh
Level doesn't exist!
there are 0 positional params remaining

$ bash test.sh -l 2
Level exist!
there are 0 positional params remaining

$ bash test.sh -l 2 foo bar
Level exist!
there are 2 positional params remaining
1   foo
2   bar

$ bash test.sh -x
test.sh: illegal option -- x

$ bash test.sh -l
test.sh: option requires an argument -- l

But you cannot put the options after the arguments: getopts stops when the first non-option argument is found

$ bash test.sh foo bar -l 2
Level doesn't exist!
there are 4 positional params remaining
1   foo
2   bar
3   -l
4   2
user_75520
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