BASH: Double quotes around part of array output?

Question

I've spent way to long trying to figure this out, so I hope someone can shed some light on this.

#!/bin/bash

HOSTNAME="`hostname`"
JSONFILE="${HOSTNAME}.json"

#####################
#     FUNCTIONS     #
#####################
function getfilesystems() {
  count=0;
  FILESYSTEMS=()
  SAVEIFS=$IFS
  IFS=$(echo -en "\n\b")
  for fs in `df -P | awk 'NR!=1'`; do
    FILESYSTEMS+=("fs$count=${fs}")
    (( count++ ))
  done
  echo "${FILESYSTEMS[@]}"
  IFS=$SAVEIFS
}

SAVEIFS=$IFS
IFS=$(echo -en "\n\b")

jo -p serverdata="$(jo hostname=${HOSTNAME} reportdata="$(date)" storage="$(jo -p "$(getfilesystems)")")"

IFS=$SAVEIFS

I'm trying to focus on this part FILESYSTEMS+=("fs$count=${fs}")

Right now, the output is this:

'fs0=/dev/mapper/vg_rpidalappnfs-LogVol00                                           10190136      5486908      4178940      57% /' 'fs1=tmpfs                                                                           1962684            4      1962680       1% /dev/shm' 'fs2=/dev/sda1                                                                        194241       104145        79856      57% /boot'

It's almost what I want. What I need is this (see the quotes difference?):

fs0="/dev/mapper/vg_rpidalappnfs-LogVol00                                           10190136      5486908      4178940      57% /" fs1="tmpfs                                                                           1962684            4      1962680       1% /dev/shm" fs2="/dev/sda1                                                                        194241       104145        79856      57% /boot"

I've literally been trying to get this to work for about an hour and I just can't seem to get past this one part.

The help is appreciated.

I don't think you want any quotes in the output; don't confuse syntactic quotes with literal quotes. You also don't really need an array in getfilesystems if all you are going to do is output the entire contents at the end; you can just write each filesystem out as you discover it. — chepner
– chepner, Commented Oct 2, 2018 at 17:05
Bash already sets the shell variable HOSTNAME for you, by the way. — Benjamin W.
– Benjamin W., Commented Oct 2, 2018 at 17:14
You can probably define the function simply as getfilesystems () { df -P | awk 'NR != 1 {print "fs"(NR-2)"="$0;}'; }. — chepner
– chepner, Commented Oct 2, 2018 at 17:35
@BenjaminW. - learned something new today; did not know that. — drewrockshard
– drewrockshard, Commented Oct 2, 2018 at 18:35

choroba · Accepted Answer · 2018-10-02 17:04:22Z

1

Escape the doublequote by a backslash:

FILESYSTEMS+=("fs$count=\"${fs}")

But I fear you won't need the doublequotes in the output in the end, but I'm not familiar with jo.

answered Oct 2, 2018 at 17:04

choroba

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2 Comments

drewrockshard Over a year ago

Already tried this before posting the questions; it did not work as expected.

drewrockshard Over a year ago

I take that back, I think this technically does what I need, but I think the jo application is messing up my output and causing it to add escapes in the shell, so I might have looked at this the wrong way when I first tried this before the question.

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