doing the following:
for i in {0..3};do
echo "${Card_${i}}"
done
my point is to get the print of parameters named "Card_1" "Card_2" and Card_3
2 Answers
It's perfectly possible to dynamically create variable names using an expansion:
$ card_1=111
$ card_2=222
$ card_3=333
$ printf '%s\n' $card_{1..3}
111
222
333
Brace expansion happens before parameter expansion, so $card_{1..3} is expanded to $card_1 $card_2 $card_3 before the parameters are expanded.
That said, it looks like you're using numerical suffixes to emulate an array:
$ cards=( 111 222 333 444 )
$ printf '%s\n' "${cards[@]:0:3}"
111
222
333
I used a slice 0:3 just to show how they work.
Comments
You could use variable indirection. However, since you've mentioned that each card is an array itself, declare -n (requires bash 4.3 or newer) might be your best option:
card_1=(c1a c1b)
card_2=(c2a c2b)
for i in {1..2}; do
declare -n arr=card_$i
echo "${arr[0]}"
done
# output:
# c1a
# c2a
Card=(1 2 3); for i in {0...3}; do echo ${Card[$i]}; done. If not, you need to use eval.eval echo "\"\${Card_$i}\"". Eval is bad.