CODEWITHSUNDEEP
bash
Carles Ràfols
Carles Ràfols

Reputation: 87

Use variable as the name of an array and get one of its elements in bash

How can I access an element of an array in bash using a variable as its name? The idea is that I want to select the array that I want by passing its name as an argument when I run my script. So I was thinking I could do something like this something like this:

#!/usr/bin/env bash

arrName=$1

g=(1 32)
g5=(5 32)
m=(1 12)
m15=(1 12)

echo "${!arrName[0]}"
echo "${!arrName[1]}"

But while this prints the first element of the array, it fails to print the second one. E.g.:

./myScript g

returns:

1

# The second line just has a new line character

Can you please explain what is going on as well as the proper way to do this (if any)?

Upvotes: 1

Views: 81

Answers (2)

chepner
chepner

Reputation: 530960

Indirection involving array elements requires a temporary variable whose value is the name of the array plus the desired index.

arrName=$1

g=(1 32)
g5=(5 32)
m=(1 12)
m15=(1 12)

tmp0="$arrName[0]"
tmp1="$arrName[1]"
echo "${!tmp0}"
echo "${!tmp1}"

That said, namerefs are easier to follow when available.

Upvotes: 1

David C. Rankin
David C. Rankin

Reputation: 84531

Indirection does not work with array names. You need to create a nameref using declare -n arrName="$1" and then remove the '!' from your echo statements, e.g.

#!/bin/bash
declare -n arrName="$1"

g=(1 32)
g5=(5 32)
m=(1 12)
m15=(1 12)

echo "${arrName[0]}"
echo "${arrName[1]}"

Example Use/Output

bash nr.sh g
1
32

Upvotes: 1

Related Questions