How do I perform double substitiution of shell variables?

Question

I have the following script stored in test.sh:

#!/bin/sh

key1=foo
key2=bar
key3=baz

for i in 1 2 3; do
    echo $i
    echo ${key${i}}
done

When I run it I get the output

1
test.sh: 9: Bad substitution

What am I doing wrong?

Answer 1

You can try

eval echo '$'key$i

Answer 2

Is this what you are looking for?

#!/bin/sh

key1=foo
key2=bar
key3=baz

for i in 1 2 3; do
  echo $i
  eval echo \$"key$i"
done

Output:

1
foo
2
bar
3
baz

Answer 3

There is an answer to this question here: https://stackoverflow.com/a/917313/10693596

Short answer is this is not possible in bash.

Update: depending on what you are ultimately after, using indirect expansion might work: https://stackoverflow.com/a/42066820/10693596