Shell Script with Options

Question

I am writing a shell script which accepts at most 2 options. However, I have a problem. Could you help me with it? (Let's say the name of this script is command.)

#!/bin/bash
PROGNAME=$(basename $0)
for OPT in "$@"
do
    echo "OPT: $OPT"
    case "$OPT" in
            '-p' | '--password' )
                    if [[ -z "$2" ]] || [[ "$2" =~ ^-+ ]]
                    then
                            echo "$PROGNAME: option requires an argument -- '$(echo $1 | sed 's/^-*//')'"
                            exit 2
                    fi
                    PASSWORD="$2"
                    shift 2
                    ;;
            '-u' | '--user' )
                    if [[ -z "$2" ]] || [[ "$2" =~ ^-+ ]]
                    then
                            echo "$PROGNAME: option requires an argument -- '$(echo $1 | sed 's/^-*//')'"
                            exit 2
                    fi
                    USER="$2"
                    shift 2
                    ;;
            * )
                    echo "$1"
                    echo "$PROGNAME: illigal option or argument -- '$(echo $1 | sed 's/^-*//')'"
                    exit 1
                    ;;
    esac
done
echo "GOOD"

Test1

$./command 
GOOD

Test2

$./command -u user
OPT: -u
OPT: user

command: illigal option or argument -- ''

(The fourth line is empty.)

Test3

$./command -u user -p passwd
OPT: -u
OPT: user
-p
command: illigal option or argument -- 'p'

Same things happens to all -u, --user, -p and --password. I have 3 questions.

1) Why does "OPT: user" show up in test2 and test3?

2) Why does this code cause errors if I give one option to it?

3) How can I fix this code? I really appreciate your help!

Why reinvent the wheel? bash has a builtin getopts

steeldriver
– steeldriver

2016-08-27 13:46:10 +00:00
Commented Aug 27, 2016 at 13:46 — steeldriver
– steeldriver, Commented Aug 27, 2016 at 13:46
Creating this code is my purpose.

Nickel
– Nickel

2016-08-27 13:52:10 +00:00
Commented Aug 27, 2016 at 13:52 — Nickel
– Nickel, Commented Aug 27, 2016 at 13:52

Stephen Harris · Accepted Answer · 2016-08-27 14:10:49Z

Your expansion of "$@" happens at the beginning of the for loop; the shift inside it won't have any impact.

eg

for a in "$@"
do
  echo OPTION: $a
  shift 2
done

With this:

$ bash x a b c d e f
OPTION: a
OPTION: b
OPTION: c
OPTION: d
OPTION: e
OPTION: f

We can see the shift isn't doing anything

Instead you want to re-evaluate it each time:

$ cat x             
while [ $# -gt 0 ]
do
  a="$1"
  echo OPTION: $a
  shift 2
done

$ bash x a b c d e f
OPTION: a
OPTION: c
OPTION: e

(You need to do a better test than this; this example code never terminates if you pass an odd number of arguments 'cos the shift 2 fails)

hschou · Accepted Answer · 2016-08-27 14:19:20Z

I use to use a while/shift loop

#! /bin/bash
PROGNAME=$(basename $0)
while [ -n "$1" ]
do
  echo "OPT: $1"
  case "$1" in
    --user=* )
      if [[ "$1" =~ --user=(.+) ]]
      then
        USER=${BASH_REMATCH[1]}
      else
        echo "$PROGNAME: option requires an argument -- $1"
        exit 2
      fi
      ;;
    -u )
      if [ -z "$2" ]
      then
        echo "$PROGNAME: option requires an argument -- $1"
        exit 2
      fi
      USER="$2"
      shift
      ;;
    * )
      echo "$PROGNAME: illigal option or argument -- $1"
      exit 1
      ;;
  esac
  shift
done
[ -n "$USER" ] && echo "USER: $USER"
echo "GOOD"

But $USER is a bad variable since it is already set.

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