bash variable array name

Question

I'm trying to let the user pick the array name because the arrays are sets of domain names.

while getopts h:c: option
do
      case "${option}"
      in
      c) client=${OPTARG};
      h) usage;;
esac
done

So like I want client=one of the customer arrays

customer1=(my_custdomain.com, my_custdomain2.com...)
custmoer2=(my1_custdomain.com, my1c_custdomain2.com...)

for i in client
do
      func_name
done

by now I'm really confused about expansions/quotes.

It isn't clear to me what you're trying to do. You seem to have: a function func_name which is passed no arguments; a list of items containing just client that you're iterating over with a variable i that is never used; it isn't clear what you are storing in $client (is it a digit, or the name of the array). I assume that custmoer2 is a typo; maybe you meant to use for i in $client. Maybe you're simply looking for the ${!var} notation for indirect expansion. Maybe you need to read the Bash manual. — Jonathan Leffler
– Jonathan Leffler, Commented Nov 11, 2015 at 4:21

chepner · Accepted Answer · 2015-11-11 12:14:49Z

2

You need indirect parameter expansion.

# No commas
customer1=(my_custdomain.com my_custdomain2.com)
customer2=(my1_custdomain.com my1c_custdomain2.com)

arrayref=$client[@]
for i in "${!arrayref}"; do
do
      func_name
done

answered Nov 11, 2015 at 12:14

chepner

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1 Comment

doc1623 Over a year ago

Perfect, Worked great. Thank U. Thank U. Thank U.

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