BASH - A variable as part of array name (part 2)

Question

Let's say I start with just item1=('item1' '1' '20')

I then define itemnumber=2

I would like to create the next array as item$itemnumber=('item2' '4' '77'), but I get a syntax error.

After that I would like to just do itemnumber=$((itemnumber+1)), and create item$itemnumber=('item3' '17' '15')

Which would give me three arrays item1, 2 and 3:

item1=('item1' '1' '20')
item2=('item2' '4' '77')
item3=('item3' '17' '15')

Is this possible?

Answer 1

A task for eval:

itemnumber=1
(( itemnumber += 1))
eval "item$itemnumber=('item$itemnumber' '4' '77')"
eval echo \${item$itemnumber[*]}