How do format and traverse an array that contains arrays ,and each array contains an array?

Question

I'm trying to create multidimensional within arrays that are contained in an array.

tests=("'0' '1 2'" "'4' '5 6'")

in each array within tests, I want to have sub arrays. With the first array "'0' '1,2'", make another for loop to go through the contents of the sub array.

Why are you trying to force bash to do this? bash only has one-dimensional arrays. Pick the right tool for the job. — glenn jackman
– glenn jackman, Commented Jun 29, 2017 at 16:32
Related on unix.SE: How can I create a multidimensional array, or something similar, with bash? - as multiple people said there, if you want this, you're better off with a programming language other than the shell, such as perl or python which can still do shell-like things like easily run other programs and capture their outputs, but also have real data structures allowing arbitrary nesting of dictionaries (hashes) and lists (arrays). — Peter Cordes
– Peter Cordes, Commented Apr 3, 2023 at 23:24

Nahuel Fouilleul · Accepted Answer · 2017-06-29 17:05:28Z

27

Since bash 4.3. (3 levels, the first contains only one element for the demo):

arr01=(0 '1 2')
arr02=(4 '5 6')
arr1=(arr01 arr02)
arr=(arr1)

declare -n elmv1 elmv2

for elmv1 in "${arr[@]}"; do
    for elmv2 in "${elmv1[@]}"; do
        for elm in "${elmv2[@]}"; do
            echo "<$elm>"
        done
    done
done

Before 4.3

arr01=(0 '1 2')
arr02=(4 '5 6')
arr1=('arr01[@]' 'arr02[@]')
arr=('arr1[@]')

for elmv1 in "${arr[@]}"; do
    for elmv2 in "${!elmv1}"; do
        for elm in "${!elmv2}"; do
            echo "<$elm>"
        done
    done
done

answered Jun 29, 2017 at 17:05

Nahuel Fouilleul

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10 Comments

clime Over a year ago

What's the point of the line declare -n elmv1 elmv2?

linux_sa Over a year ago

@clime, if you have bash 4.3 or later, type man bash and search for declare -n.

kebs Over a year ago

@linux_sa Sorry, still unclear. My bash 4.3.48 man says:...running declare -n ref=$1 inside the function creates ..., but the answer here states declare -n elmv1 elmv2 without the =. Is that the same? Quite unclear to me. And help declare isn't much help either: -n make NAME a reference to the variable named by its value. That is pretty much meaningless to me.

Nahuel Fouilleul Over a year ago

@Timo, This answer seems to explain this last sentence.

Nahuel Fouilleul Over a year ago

a=( 1 2 ); declare -n a=b; --> bash: declare: a: reference variable cannot be an array otherwise declare -n b=a; b=(3 4); declare -p a b

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