Exclude getopts-options from being read into array by $@

user_15051374 0
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I'm trying to make a script with the following usage:

bash script.sh [-a] databases ...

The script then populates the array $databases with the arguments from 'databases'

databases=("$@")

Now I'm trying to implement the option -a which if set should declare att=1:

while getopts "a" o; do
    case "${o}" in
        a)
            att=1
            ;;
    esac
done

Now I'm facing the problem that when executing the script -a also gets put into the array.

I can't use read for the array because I don't have the option to enter additional arguments after the script gets executed as the script will be automated with a RMM and that only lets you enter addition arguments when creating a job.

scripting arrays bash
2 Answers

You have to use shift in case block before you populate your variable:

while getopts "a" o; do
   case $o in
      a)
         att=1
         shift
         ;;
    esac
done

databases=("$@")
declare -p databases att

Then run it like this:

bash script.sh -a DB1 DB2 DB3

Output:

declare -a databases=([0]="DB1" [1]="DB2" [2]="DB3")
declare -- att="1"
user_5482250

The standard way to manage options with getopts is to do a global shift :

#!/usr/bin/env bash

while getopts "a" o; do
    case "${o}" in
    a) att=1 ;;
    esac
done
shift $((OPTIND - 1))

databases=("$@")

declare -p databases att
# output
# ./test.sh -a DB1 DB2 DB3
# declare -a databases=([0]="DB1" [1]="DB2" [2]="DB3")
# declare -- att="1"
user_21256710
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