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I am trying to print the values of a variable which has naming sequentially in bash.

boxes1=abcd boxes2=efgh boxes3=ijkl

for (( i=1; i<=$boxCount; i++ )); 
   do
     echo "\$$boxes{i}"
   done

I want output like: abcd efgh ijkl

ilkkachu
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sandeep yashodha Shiviraju
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  • Using a nameref [Variable from concatentation of variables](https://unix.stackexchange.com/questions/337113/variable-from-concatentation-of-variables/337117#337117) or an indirect reference [Use a variable reference “inside” another variable](https://unix.stackexchange.com/questions/41406/use-a-variable-reference-inside-another-variable/41418#41418) – steeldriver Aug 04 '21 at 16:56
  • For indexed variables, use arrays: https://mywiki.wooledge.org/BashGuide/Arrays For other cases where you have a variable name in another variable, namerefs (`declare -n`) or indirect refs (`${!p}`) – ilkkachu Aug 04 '21 at 16:58
  • I tried the above mentioned options, but it didn't work. Could you help further pls. – sandeep yashodha Shiviraju Aug 04 '21 at 21:35
  • Add the exact command you tried, and any error messages, to your question. Alternatively, ask a new question about the use of arrays, namerefs or indirect variables. Using a sequence of distinct names is not viable: every time you add another name in the series, you need to update (and test) significant amounts of code. See "Two or more, use a for", or possibly DRT (Don't repeat yourself). – Paul_Pedant Aug 04 '21 at 21:42
  • Below is the command and the error: for(( i=1; i<=3; i++ )); do declare -n var="boxes$i"; printf 'boxes%d = %s\n' "$i" "$var"; done -bash: declare: -n: invalid option declare: usage: declare [-afFirtx] [-p] [name[=value] ...] boxes1 = -bash: declare: -n: invalid option declare: usage: declare [-afFirtx] [-p] [name[=value] ...] boxes2 = -bash: declare: -n: invalid option declare: usage: declare [-afFirtx] [-p] [name[=value] ...] boxes3 = – sandeep yashodha Shiviraju Aug 04 '21 at 22:32
  • @sandeepyashodhaShiviraju You need to define `boxes1 boxes2 boxes3` before you declare a nameref that references them. Also `for (( i=1; i<=3; i++ )); do var=boxes$i; echo "${!var}"; done`. – Paul_Pedant Aug 05 '21 at 23:13

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