Filling a dynamic array with user input in shell scripting

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?

You can get rid of $index and eval. Simply replace eval Name[$index]= $io with Name=("${Name[@]}" "$io"). — alvits
– alvits, Commented Jun 27, 2016 at 19:30

tony19 · Accepted Answer · 2016-06-27 19:18:50Z

3

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

answered Jun 27, 2016 at 19:18

tony19

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1 Comment

S_Ananth Over a year ago

Thanks @tony19.It worked for me.I am trying to print multiple array values inside the same loop, for example: person: 1 - print name,age, dob,etc ..so i followed the same technique of declaring 3 arrays. When i tried to do " echo person 1 {$Name[$i]}" , i am not able to access the corresponding value of the variable Name. rather it just prints "Name[0]" . Is there any syntax for this case?

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