How to read command line argument in shell sript

Question

I need to read command line argument which is passing as scriptname –c "30,31,32,33,34,35"

and convert it to

myArray=( 30 31 32 )

in the argument any number between 31,32,33,34,35 can be passed as –c "30,31,32,33,34,35" — pri, Jun 12 '15 at 11:38
And why are 33, 34, and 35 in the example given at command line but omitted in the array? That's confusing. — BlackJack, Jun 16 '15 at 13:33

score 1 · Answer 1 · edited May 23 '17 at 12:05

1

You can use getopts command to read the arguments. Please refer the link for the usage example of how to use getopts in bash

Once you have the variables, you can easily create an array This link might be useful. Arrays in unix shell?

edited May 23 '17 at 12:05

Community

1
1

answered Jun 11 '15 at 11:01

sharathkj

11
3

score 1 · Accepted Answer · answered Jun 11 '15 at 13:02

Try the following:

while getopts c: option
do
    case $option in
       c) data="$OPTARG"
          ;;
    esac
done

oldIFS="$IFS"
IFS=','
myArray=($data)
IFS="$oldIFS"

echo ${myArray[@]}

The c: after getoptsindicates that we have an option -c, the : indicates it is followed by an argument, which is retrieved using $OPTARG.

IFS if the Inter Field Separator which I reset to a comma in order to create the array.

How to read command line argument in shell sript

2 Answers2