How get next to next argument in linux shell script?

Question

I have wrote simple shell script test.sh as follows:

while getopts ":A:" OPTION
do 
   case $OPTION in
   A)
      echo $OPTARG
   ?)
      echo "no option"

  esac
done

And executed the scripts as follows

$ ./test.sh -A 1 2

Now if got argument 1 by $OPTARG but how can i access the second argument ( 2 in this case)?

Regards Jayesh

Can you elaborate in little more detail like what exactly you want to do with above script? — Yogesh, Jan 02 '14 at 12:06
okay if I understand you correctly you are trying to bind both 1 and 2 with `-A` option, you can pass 2 with another switch like `./test.sh -A 1 -B 2` and then parse them. Here is a helpful link => http://stackoverflow.com/questions/16483119/example-of-how-to-use-getopt-in-bash — Ashish Gaur, Jan 02 '14 at 12:11

Igor Chubin · Accepted Answer · 2014-01-02T12:47:38.197

3

There are several options.

(1) You can use shift and take $1

 while -n "$1"
 do  
  # do something with $1
  shift
 done

(2) You can iterate through the args:

 for i
 do
    # do something with $i
 done

There are other alternatives also.

edited Jan 02 '14 at 12:47

answered Jan 02 '14 at 12:06

Igor Chubin

1 Answers1