How to print out name of array variable in bash script

Question

I have a bash script like:

#!/bin/bash

tom[0]="28 years old, computer science bs"
tom[1]="41 years old, physics bs"

ryan[0]="32 years old, technician"

mary[0]="25 years old, math bs"
mary[1]="30 years old, chemistry ba"

robert[0]="34 years old, math bs"

for name in robert mary tom ryan  
do  
    echo "name = ${name}:" 
    for who in "${name[@]}"
    do  
        echo "age,degree = ${!name}"  
    done
done

exit 0

output:

name = robert:
age,degree = 34 years old, math bs
name = mary:
age,degree = 25 years old, math bs
name = tom:
age,degree = 28 years old, computer science bs
name = ryan:
age,degree = 32 years old, technician

How do I modify my bash script to display below output:

name = robert:
age,degree = 34 years old, math bs

name = mary:
age,degree = 25 years old, math bs
age,degree = 30 years old, chemistry ba

name = tom:
age,degree = 28 years old, computer science bs
age,degree = 41 years old, physics bs

name = ryan:
age,degree = 32 years old, technician

It's not clear what you mean. Just do echo mary or echo tom. — Barmar
– Barmar, Commented May 30, 2023 at 20:00
It seems like you probably want an associative array, not multiple array variables. info['tom0']="28 years old, computer science bs"; info['mary0']="25 years old, math bs"; info[mary1]="30 years old, chemistry ba"; — Barmar
– Barmar, Commented May 30, 2023 at 20:04
Unfortunately, bash doesn't have 2-dimensional arrays. It's probably not the best language to use for an application like this. — Barmar
– Barmar, Commented May 30, 2023 at 20:06
It sounds very much like you have the wrong approach to whatever you're trying to do. If you ask a question about THAT, including a minimal reproducible example with concise, testable sample input, expected output, and code, then we can help you better than showing you how to keep going down your current path. — Ed Morton
– Ed Morton, Commented May 30, 2023 at 20:56
@novi2023 we all understood what you wanted to do, it wasn't confusing, but we wanted to help you come up with a better approach as you seemed to be asking an XY Question. — Ed Morton
– Ed Morton, Commented May 30, 2023 at 22:23

Andrej Podzimek · Accepted Answer · 2023-05-31 13:18:17Z

0

Bash arrays are tricky. While "${!name}" gives you access to a scalar variable by name (or to an array’s zeroth element), you cannot access array elements other than [0] directly this way. For example, the construct "${!name[@]}" would expand to the array’s indices rather than to the stored values.

BTW, "${name[@]}" makes little sense in your code; it is equivalent to "$name" in this special case, because a scalar string "$name" is also (implicitly) "${name[0]}" which (in absence of other array elements) equals "${name[@]}".

To access Bash arrays by name, you need almost-pointers in the form of declare -n. That gives you an alias which behaves (almost) like the referenced variable (array in this case):

tom=('28 years old, computer science bs'
     '41 years old, physics bs')
ryan=('32 years old, technician')
mary=('25 years old, math bs'
      '30 years old, chemistry ba')
robert=('34 years old, math bs')

for name in robert mary tom ryan; do
  printf '\nname = %s:\n' "$name"
  declare -n array="$name"
  printf '%s\n' "${array[@]}"
done

In the snippet above one could also loop over "${array[@]}" using a for-loop, but one can just as well let printf handle a few extra arguments.

answered May 31, 2023 at 13:18

Andrej Podzimek

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2 Comments

novi2023 Over a year ago

Thank you @Andrej for spending time to explain the differences of how to access the bash arrays. I appreciate it. I can use your solutions, I just have to re-arrange the data ( index arrays) passed down to me. I still would prefer not to alter the original data, but if I can't find any other way, then I will re-arrange those given data.

novi2023 Over a year ago

Your solutions actually worked. After I played around with it, I didn't have to modify any of the arrays given at all. I used for loop and it actually displayed what I was trying to accomplish. Thanks a lot.

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