bash getopts uses option/flag as argument

Question

I'm writing a bash shell script that can use the flags -l -u and -w [argument]

I have the following (simplified) code:

while getopts ":ulw:" arg; do
    case "$arg" in
    u)
        U=1
        ;;
    l)
        L=1
        ;;

    w)
        W=1
        VALUE="${OPTARG}"
        ;;

    esac
done

when I run my script with -w42 -l it works like it should. If I use -lw42 it also works but when I use -w42l, it thinks 42l is the argument (instead of just 42) and it makes the VALUE variable = 42l and ignores the -l option.

How can I make the script work for both -w42 -l, -lw42 and -w42l?

EDIT: to clear things up, I know this isn't the way it's supposed to work but my college requires me to make it work this way

`-w42l` is working exactly the way it's supposed to. Your expectation that it would be like `-w 42 -l` is wrong -- I've never seen a program where that expectation would be met; it's not the standard way to do things. — Charles Duffy, Dec 19 '21 at 17:05
See the "Utility Syntax Guidelines" section at the end of https://pubs.opengroup.org/onlinepubs/9699919799.2016edition/basedefs/V1_chap12.html describing how programs are supposed to parse command-line arguments. In that context, `42` is what's called an "option argument". — Charles Duffy, Dec 19 '21 at 17:06
Tangentially see also https://stackoverflow.com/questions/673055/correct-bash-and-shell-script-variable-capitalization — tripleee, Dec 19 '21 at 17:08
Ah! Okay, if your college is forcing it that's different. Do you really need to use `getopts`, or can you implement your own parser? — Charles Duffy, Dec 19 '21 at 17:11
Why isn't `-w42l` the same as `-w 42l`? What should happen on `-wanything_l_another_l_thing`? — KamilCuk, Dec 19 '21 at 17:22
@Mr_BananaPants, ...btw, you might want to read [BashFAQ #35](http://mywiki.wooledge.org/BashFAQ/035) for a general introduction to option-parsing in bash. As it says, getopts covers the simple cases, but it's often inadequate for more complex needs. — Charles Duffy, Dec 19 '21 at 17:40
@KamilCuk I believe the OP expects that `-w42l` is similar to `-w 42 -l` — kvantour, Dec 19 '21 at 21:37

Charles Duffy · Accepted Answer · 2021-12-19T17:41:24.720

On Standards-Compliance

What you are trying is not supposed to work in the first place.

POSIX utility syntax guideline #5 states:

One or more options without option-arguments, followed by at most one option that takes an option-argument, should be accepted when grouped behind one '-' delimiter.

So, the option taking an option-argument (-w in this case) is only allowed to be the last one in a group started by a single -.

On Making It Work Anyhow

If you can't deal with standard-compliant behavior, you can't use getopts, so you need to write your own logic. One way to do that might look like the following:

#!/usr/bin/env bash
#              ^^^^- note bash, not sh; the below code uses non-POSIX extensions

while (( $# )) && [[ $1 = -* ]]; do
  arg=${1#-}; shift
  while [[ $arg ]]; do
    case $arg in
      l*) flag_l=1; arg=${arg#l};;
      u*) flag_u=1; arg=${arg#u};;
      w*)
        flag_w=1
        rest=${arg#w}
        if [[ -z $rest ]]; then
          arg=$1; shift; rest=$arg
        fi
        if [[ $rest =~ ^([[:digit:]]+)(.*) ]]; then
          w_value=${BASH_REMATCH[1]}
          arg=${BASH_REMATCH[2]}
        else
          echo "ERROR: -w not followed with a number" >&2
          exit 1
        fi
        ;;
      *) echo "Unrecognized flag: $arg" >&2; exit 1;;
    esac
  done
done

echo "After parsing:"
echo "flag_w = ${flag_w:-0}"
echo "flag_l = ${flag_l:-0}"
echo "flag_u = ${flag_u:-0}"
echo "w_value = ${w_value:-0}"

See this running in the online interpreter at https://ideone.com/eDrlHd

bash getopts uses option/flag as argument

1 Answers1

On Standards-Compliance

On Making It Work Anyhow