I have this little piece of code:
#!/bin/bash
item01=('item1' '1' '20')
item02=('item2' '4' '77')
item03=('item3' '17' '15')
zeroone=01
zerotwo=02
echo ""
declare -a array=()
array=${item$zeroone[@]}
echo ""
echo ${array[@]}
echo ""
Obviously this doesn't work (bad substitution).
Is there a way to make it work? Such that a variable can be a part of an array name?
And also, to make this work in particular:
array[0]=${item$zeroone[0]}
and
another_variable=${item$zeroone[0]}
Thx
Better use associative arrays:
declare -A item=([1, 0]='item1' [1, 1]='1' [1, 2]='20')
...
Accessing an element:
one=1
echo "${item[$one, 0]}"
On a loop:
for ((I = 0; I <= 2; ++I)); do
echo "${item[$one, $i]}"
done
You can also use strings instead of numbers:
declare -A item=(["01", 0]='item1' ["01", 1]='1' ["01", 2]='20')
Another answer: You can actually use references:
item01=('item1' '1' '20')
item02=('item2' '4' '77')
item03=('item3' '17' '15')
zeroone=01
zerotwo=02
echo ""
ref="item${zeroone}[@]"
declare -a array=("${!ref}") ## Still produces 3 arguments as if "${item01[@]}" was called
echo ""
echo "${array[@]}"
echo ""
