Substituting bash for loop variable to get array name

Question

I'm trying to use a bash for loop to choose an array, the members of which I want to use as variables in a for loop one level down. My issue is in doing the expansion in the second for loop, which I can't figure out.

first_two_models=( model_one model_two )
second_two_models=( model_three model_four )


for name in first_two second_two; do
    for model in ${"$name_models"[@]}; do
        echo $model
    done
done

Thank you for any help you can offer!

What you're looking to do (for purposes of easier searching) is called "indirect array expansion". See [BashFAQ #6](http://mywiki.wooledge.org/BashFAQ/006#Evaluating_indirect.2Freference_variables) -- note that the "nameref" syntax discussed there is available in bash 4.3. — Charles Duffy, Jun 02 '17 at 15:23

Nahuel Fouilleul · Answer 1 · 2017-06-02T15:29:08.513

3

for name in 'first_two_models[@]' 'second_two_models[@]'; do
    for model in "${!name}"; do
        echo "$model"
    done
done

edited Jun 02 '17 at 15:29

answered Jun 02 '17 at 15:09

Nahuel Fouilleul

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Try running this after `second_two_models+=( "* Model Four *" )` – Charles Duffy Jun 02 '17 at 15:22
I understand your point is about `echo $model` without the quotes, I didn't focus on this point, but I will update – Nahuel Fouilleul Jun 02 '17 at 15:29

bishop · Accepted Answer · 2017-06-02T15:40:24.580

2

You need to first build the array reference as a string, then use indirect reference on that string to pull the values:

first_two_models=( model_one model_two )
second_two_models=( model_three model_four )

for name in first_two second_two; do
    array="${name}_models[@]"     # build reference as string
    for model in "${!array}"; do  # use indirect reference to access
        echo "$model"
    done
done

edited Jun 02 '17 at 15:40

answered Jun 02 '17 at 15:10

bishop

37,830
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Might consider fixing up the code so it still works if `second_two_models+=( "Model Five" )`; right now, `Model` and `Five` are separate iterations for the inner array. – Charles Duffy Jun 02 '17 at 15:21
Thanks @CharlesDuffy! Fixed. – bishop Jun 02 '17 at 15:28

Substituting bash for loop variable to get array name

2 Answers2