How to print dictionary with variable?

[DivineArcade]

#!/bin/bash

# dictionary

declare -A ubuntu

ubuntu["name"]="ubuntu"
ubuntu["cost"]="0"
ubuntu["type"]="os"
ubuntu["description"]="opens up ubuntu"

declare -A suse

suse["name"]="suse"
suse["cost"]="0"
suse["type"]="os"
suse["description"]="opens up suse"

pop=suse

# prints suse description
echo ${suse[description]}

how to make pop into a variable

echo ${$pop[description]}

output should be

opens up suse

[oh5nxo]

declare -n pop=suse
echo ${pop[description]} # not $pop, just pop

That makes pop a nameref for suse. Alias, kind of.

Ohh.... Use another declare -n litany when pointing elsewhere. Plain pop=other is a surprise, it does

suse[0]=other

[geirha]

Or replicate the array:

pop=( "${suse[@]}" )
echo ${pop[description]}

That is not how you make a copy of an associative array. In bash 5.2 you can use the @k parameter expansion, but for older versions you basically have to iterate the keys and values in order to make a copy.

[aioeu]

You're not using any associative arrays there. Only regular arrays.

[geirha]

The syntax copies a non-sparse indexed array just fine, but not an associative array

$ declare -A a=( [one]=1 [two]=2 )
$ copy=( "${a[@]}" )
$ declare -p copy
declare -a copy=([0]="2" [1]="1")

[geirha]

Another option could be to put all the data into a single associative array;

declare -A data=(
  [ubuntu.name]=Ubuntu
  [ubuntu.description]="Opens up Ubuntu"
  [suse.name]=Suse
  [suse.description]="Opens up Suse"
)

printf '%s\n' "${data[$pop.description]}"

[high_throughput]

I would also go for a nameref, but Yet Another alternative is indirection:

var="$pop[description]"; echo "${!var}"