use variable on array name in for loop in bash

Question

I want to print out the element in a list:

lst_1=(A B C)
for j in ${lst_1[@]};do echo $j ;done
A
B
C

this worked. But when I use variable on list name, it failed.

lst_1=(A B C)
lst_2=(D E F)

for i in 1 2;do \
    for j in ${lst_${i}[@]};do \
        echo $j ;done;done

-bash: ${lst_${i}[@]}: bad substitution

`for l in lst_{1,2}'[@]'; do for e in "${!l}"; do echo "$e"; done; done` — Fravadona, Jun 17 '22 at 12:40
Thanks! Could you explain what does the "!" means in "${!l}" ? — mike, Jun 17 '22 at 12:42
if `var="lst_1[@]"` then `"${!var}"` is like expanding `"${lst_1[@]}"` — Fravadona, Jun 17 '22 at 12:47

score 1 · Accepted Answer · answered Jun 17 '22 at 11:23

1

This will work:

lst_1=(A B C)
lst_2=(D E F)

for i in 1 2; do
    for j in $(eval echo \${lst_${i}[@]}); do
        echo $j ;
    done
done

This make the variable evaluations in 2 steps: first ${i} is substituted. Then the result (which will look like "echo ${lst_1[@]}" because the first dollar was escaped) will be interpreted by the eval command, resulting in the expanded array.

Note: not sure this solution will work in all cases (for example if the array elements will have spaces in them), other solutions might be better.

answered Jun 17 '22 at 11:23

vladmihaisima

2,119
16
20

The use of `eval` is not required: `ref=lst_$i[@]; for j in "${!ref}"; ...` – Fravadona Jun 17 '22 at 12:00
Different solutions require different constructs (the above solution eval, the indirection a separate variable). – vladmihaisima Jun 17 '22 at 12:14
@vladmihaisima thanks! this is exact what I want, I guessed the eval may be useful, but not familiar with it. – mike Jun 17 '22 at 12:40

use variable on array name in for loop in bash

1 Answers1