How to write user defined options in shell script?

Question

Possible Duplicate:
Using getopts in bash shell script to get long and short command line options

This is the code i've written

#! /bin/bash

getopts master $1

while getopts ":master" opt; do
  case $opt in
    master)
      echo "-master was triggered! $1 was entered" >&2
      ;;
    \?)
      echo "Invalid option: -$OPTARG" >&2
      ;;
  esac
done

and this is the output i'm getting-

]$ ./test123.sh -master 123
./test123.sh: line 3: getopts: `-master': not a valid identifier

How do i define a user defined option?

score 3 · Accepted Answer · answered Dec 06 '12 at 11:55

3

I think there are a couple of problems.

I don't understand what this (getopts master $1) is doing and you can remove it
I think you need single char arguments (i.e. -m vs -master)

e.g. the below seems to work:

#!/bin/bash
while getopts ":m" opt; do
  case $opt in
    m)
      echo "-m was triggered! $1 was entered" >&2
      ;;
    \?)
      echo "Invalid option: -$OPTARG" >&2
      ;;
  esac
done

answered Dec 06 '12 at 11:55

Brian Agnew

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when i run the script the output i get is `-m was triggered! -m was entered`, but i want the output to be `-m was triggered! 123 was entered` – AlwaysALearner Dec 06 '12 at 12:00
got it. i must writ it as `echo "-m was triggered! $2 was entered" >&2` – AlwaysALearner Dec 06 '12 at 12:05

How to write user defined options in shell script?

1 Answers1