Filling a dynamic array with user input in shell scripting - arrays

Question

I am new to shell scripting. I am trying to create an array of size n, where n is input by the user during the run time.
while [ $i -lt $n ]
do
echo For person $i enter the name?
read io
eval Name[$index]= $io
done
When I try to do this, the values are overwritten every time the loop gets the input from user.
For ex: if person 1 is - Tom,if person 2 is - John. Then when i try to print the names of all person at the end of the script, person 1 name is overwritten with person n th name.(which means, all names are stored in a single variable instead of an array).
Can someone tell me where am i going wrong?

Answer 1

You need to increment i in the loop so that it eventually exits. This line increments i by 1:
let i+=1
You don't need to use eval in eval Name[$index]= $io.
There is no variable named index (at least not in your code sample). I assume you meant to use i there. (i.e., Name[$index] should be Name[$i])
This code works:
#!/bin/sh -e
Name=()
i=0
while [ $i -lt 4 ]
do
echo For person $i enter the name?
read io
Name[$i]=${io}
let i+=1
done
echo names:
for n in "${Name[#]}"
do
echo $n
done