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I am having unexpected results in expanding an array. I am hoping someone can shed some light on why I am seeing what I am seeing. I am trying to populate an array using fdisk -l to fill it and only getting a single element. The code:

#!/bin/bash
declare -a PARTITIONS=();

PARTITIONS=$(fdisk -l ubuntu-minimal-16.04-desktop-armhf-raspberry-pi-2.img | grep -i ubuntu-minimal-16.04-desktop-armhf-raspberry-pi-2.img | sed '/Disk/d' | cut -d " " -f1)

echo "PARTITIONS[@]:${PARTITIONS[@]}"

echo "ELEMENT 0: ${PARTITIONS[0]}"

echo "ELEMENT 1: ${PARTITIONS[1]}"

Output:

 PARTITIONS[@]:ubuntu-minimal-16.04-desktop-armhf-raspberry-pi-2.img1 ubuntu-minimal-16.04-desktop-armhf-raspberry-pi-2.img2

 ELEMENT 0: ubuntu-minimal-16.04-desktop-armhf-raspberry-pi-2.img1 ubuntu-minimal-16.04-desktop-armhf-raspberry-pi-2.img2

 ELEMENT 1:

There is no element 1. What am I doing wrong?

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  • 1
    What are you expecting to be in element 1? If you're expecting it to be ubuntu-minimal-16.04-desktop-armhf-raspberry-pi-2.img2, you may notice that it is part of element 0. Commented Jul 15, 2016 at 18:11

2 Answers 2

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When assigning to an array:

array=( elements )

I.e.,

PARTITIONS=( $(fdisk ... ) )
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When a string is assigned to an array, it is assigned to its first element. Observe:

$ declare -a x
$ x=$(date)
$ declare -p x
declare -a x='([0]="Fri Jul 15 11:09:59 PDT 2016")'

Here $(date) is the string that results from command substitution. (This is like your fdisk pipeline but simpler.) declare -p is used here to show the exact contents of a bash variable.

If you want word splitting on the string, use parens like so:

$ x=($(date))
$ declare -p x
declare -a x='([0]="Fri" [1]="Jul" [2]="15" [3]="11:10:08" [4]="PDT" [5]="2016")'
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