How to set some variable to some value when option is present

Question

I am not good at Bash scripting, and trying to learn more. Let me introduce my question with code:

#!/bin/bash
version_num=
isWindows=
MSVC_VER=
VERBOSE=
while getopts “hw:v:m:V” OPTION
do
     case $OPTION in
         h)
             usage
             exit 1
             ;;
         w)
             isWindows=$OPTARG
             ;;
         v)
             version_num=$OPTARG
             ;;
         m)
             MSVC_VER=$OPTARG
             ;;
         V)
             VERBOSE=1
             ;;
         ?)
             usage
             exit
             ;;
     esac
done

For space, usage function is removed.

My questions are:

First question:

currently, if I use this script, I have to feed parameter values after each option, for example:

 `bash test_bash.sh -v 4.2.2.0 -m 10.0 -w 1`

However, assuming that I only need to see whether -w is present, then set some variable value. I don't want to provide -w 1 since 1 is just a flag to do something. I would like the script to work like:

 bash test_bash.sh -w -v 4.2.2.0 -m 10.0

How can I achieve this? I would like to do something like rm -r -f, which can have multiple options and does not require that each option is followed by some value.

Second question:

if I remove

V)
   VERBOSE=1 
   ;;

and :V from the while line as well as VERBOSE=, this script does not work anymore. Is it because :V is required?

Thanks a lot for your time and help!

Barmar · Accepted Answer · 2013-05-07 17:40:48Z

4

Putting a : after a letter in the getopts parameter indicates whether it takes a parameter after it or not. So change to:

while getopts “hwv:m:V” OPTION

Removeing :V from the script breaks it because the : is for the m option that comes before it, not the V option that comes after. When you remove that :, it means that m no longer takes a parameter, but you need that.

answered May 7, 2013 at 17:40

Barmar

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