How can I populate an associative array with a variable name in bash?

Question

I am trying to create an associative array with a variable name and having some trouble. I am able to declare the array without any trouble, but when I try to populate it I get an error.

Here is my code:

ARR_NAME="tali"

makeArray () {
  name=${ARR_NAME}_$1
  echo name: $name
  declare -A $name  #this works fine
  ${name}=( [foo]=bar [baz]=quux [corge]=grault )  #this gives an error (see below)
}

makeArray dev

Here are the error messages I get:

./array_test.sh: line 135: syntax error near unexpected token `[one]=uno'
./array_test.sh: line 135: `  ${name}=( [one]=uno [two]=dos [three]=tres )'

Other things I have tried:

$name=( [one]=uno [two]=dos [three]=tres ) #same error message as above

declare -A $name=( [one]=uno [two]=dos [three]=tres ) 
# gives me the following error (basically the same as above):
./array_test.sh: line 134: syntax error near unexpected token `('
./array_test.sh: line 134: `  declare -A $name=( [one]=uno [two]=dos [three]=tres )'

I am new to bash and have been Googling anything and everything possible to figure this out, but no luck so far. The closest answer I have found is this, but it didn't solve my problem. I am using bash 4.3.

Note: this is just a practice script I am using to learn more about bash and associative arrays. I am testing things out here to use in the actual script I am trying to write.

That linked answer should solve your problem. You need a nameref here. What about it didn't work? What did you try exactly? To clarify you won't be able to use the variable directly but you can give a local alias to the name in the variable. — Etan Reisner
– Etan Reisner, Commented Dec 30, 2015 at 20:12
declare -A $name="( [one]=uno [two]=dos [three]=tres )" will declare and define the array. — rici
– rici, Commented Dec 30, 2015 at 20:22

Armali · Accepted Answer · 2018-02-12 13:27:41Z

0

declare -A $name="( [one]=uno [two]=dos [three]=tres )" will declare and define the array. – rici

This is true, but missing the important -g option to make the array accessible outside of makeArray:

The -g option forces variables to be created or modified at the global scope, even when declare is executed in a shell function.

So it's

  declare -gA $name="( [foo]=bar [baz]=quux [corge]=grault )"

answered Feb 12, 2018 at 13:27

Armali

19.6k15 gold badges64 silver badges184 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

How can I populate an associative array with a variable name in bash?

1 Answer 1

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related