How do you reference a variable within another variable in bash? - arrays

Question

I am trying to do something which I thought would be fairly simple but not having any joy.
If I run the following, it successfully pulls the region (reg) from an array called PORT330 and checks to see it contains the value of $i. For example, it could be checking to see if "Europe London" contains the word "London". Here is what works:
if [[ ${PORT330[reg]} == *"$i"* ]] ; then
echo 302 is in $i ;
fi
However, I actually have a list of Port array's to check so it may be PORT330, PORT550 and so on. I want to be able to substitute the port number with a variable but then call it within a variable. Here is what I am trying to do:
This works:
for portid in ${portids[#]} ; do
for i in ${regions[#]} ; do
if [[ ${PORT330[reg]} == *"$i"* ]] ; then
echo $portid is in $i ;
fi ;
done ;
done
However this doesn't work:
for portid in ${portids[#]} ; do
for i in ${regions[#]} ; do
if [[ ${PORT$portid[reg]} == *"$i"* ]] ; then
echo $portid is in $i ;
fi ;
done ;
done
It throws this error:
-su: ${PORT$portid[reg]}: bad substitution
Any pointers as to where I am going wrong?

Answer 1

In BASH variable expansion there is an option to have indirection via the ${!name} and ${!name[index]} schemes
for portid in ${portids[#]} ; do
for i in ${regions[#]} ; do
arr=PORT$portid
if [[ ${!arr[reg]} == *"$i"* ]] ; then
echo $portid is in $i ;
fi ;
done ;
done
Here is a complete example
PORT330[reg]=a ;
PORT550[reg]=b ;
for portid in 330 550 ; do
for i in a b ;
do arr=PORT$portid ;
if [[ ${!arr[reg]} == *"$i"* ]] ; then
echo $portid is in $i ;
fi ;
done ;
done
Produces
330 is in a
550 is in b
Another example
:~> portids=(330 350 )
:~> echo ${portids[#]}
330 350
:~> PORT350[reg]=London
:~> PORT330[reg]=Berlin
:~> for portid in ${portids[#]} ; do arr=PORT$portid ; echo ${!arr[reg]} ; done
Berlin
London