BASH array variable name concatenating

Question

I’ve got a little problem with the variable name concatenation oder replacement in BASH. Given is:

#!/bin/bash
Audio1=(0:ita, 96k, AAC, 2.0, s16le, 48000)   # audio stream definitions
Audio2=(1:rus, 128k, AAC, 2.0, s16le, 48000)  # for use in ffmpeg or
Audio3=(2:klg, 96k, AAC, 1.0, s16le, 48000)   # avconv
# and so on, now processing the audio streams
for ((i=1 ; i<=8 ; i++)) ; do
  Audio="Audio$i"
  Audio=${!Audio}
  if [ ${#Audio[0]} -gt 0 ] ; then
    # do the job with $Audio, here examplaryly:
    echo ${Audio[@]}
  fi
done

Expected as result was:

0:ita, 96k, AAC, 2.0, s16le, 48000
1:rus, 128k, AAC, 2.0, s16le, 48000
2:klg, 96k, AAC, 1.0, s16le, 48000

I got:

0:ita,
1:rus,
2:klg,

It’s only half the solution, because there is only the first element copied of each source array instead of been totally copied into the working array $Audio. In PHP I would write the variable name replacement simply by:

for ($i=1; $i<=8; $i++) {
  $Audio = ${'Audio'.$i};     // ← that’s all the magic in PHP ;-)
  if (isset($Audio[0])) {
    // blabla
  }
}

Unfortunatelly, I can’t use PHP for this application. Of course, this is one of my first scripts in BASH. Switching over to any other coding (e.g. many simple variables instead of few arrays) aren’t an option. I have to use these presets. So, where’s my fault, what’s going wrong?

Greetings - Bert

Array elements are separated by whitespace, not commas.

chepner
– chepner

2016-03-11 21:20:13 +00:00
Commented Mar 11, 2016 at 21:20 — chepner
– chepner, Commented Mar 11, 2016 at 21:20
Thanks. I feel, that will be not the last mistake.

Bert
– Bert

2016-03-11 21:36:26 +00:00
Commented Mar 11, 2016 at 21:36 — Bert
– Bert, Commented Mar 11, 2016 at 21:36

anubhava · Accepted Answer · 2016-03-11 21:26:28Z

1

In BASH versions 4.3+ you can use declare -n:

for ((i=1 ; i<=3; i++)) ; do
   declare -n Audio="Audio$i"
   if [[ ${#Audio[@]} -gt 0 ]]; then
       echo "${Audio[@]}"
   fi
done

0:ita, 96k, AAC, 2.0, s16le, 48000
1:rus, 128k, AAC, 2.0, s16le, 48000
2:klg, 96k, AAC, 1.0, s16le, 48000

-n makes NAME a reference to the variable named by its value. So in above code Audio is a reference to the value contained in Audio1, Audio2, Audio3 etc.

edited Mar 11, 2016 at 21:26

answered Mar 11, 2016 at 20:25

anubhava

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5 Comments

Bert Over a year ago

Thanks, anubhava! I think, it’s the best solution. I read many explanations about the variable name manipulation, but this item "declare" I have jumped over everytime. It seemed not to be the right thing.

anubhava Over a year ago

-n makes NAME a reference to the variable named by its value. So in above code Audio is a reference to the value contained in $Audio

anubhava Over a year ago

Thanks @chepner, yes indeed it is 4.3+

glenn jackman Over a year ago

@Bert, if you have the right bash version, declare -n is exactly what you want.

Bert Over a year ago

bash says:GNU bash, Version 4.3.30(1)-release-(i586-pc-linux-gnu)

glenn jackman · Accepted Answer · 2016-03-11 21:55:51Z

1

Without using namerefs, you can do:

Audio1=(0:ita, 96k, AAC, 2.0, s16le, 48000)   # audio stream definitions
Audio2=(1:rus, 128k, AAC, 2.0, s16le, 48000)  # for use in ffmpeg or
Audio3=(2:klg, 96k, AAC, 1.0, s16le, 48000)   # avconv

for i in {1..3}; do
    aryvar="Audio${i}[@]"   # this is the special magic indirect varname
    for value in "${!aryvar}"; do printf "%d\t%s\n" $i "$value"; done
    # ...........^^^^^^^^^^^^ expands to elements of the array
done

outputs

1   0:ita,
1   96k,
1   AAC,
1   2.0,
1   s16le,
1   48000
2   1:rus,
2   128k,
2   AAC,
2   2.0,
2   s16le,
2   48000
3   2:klg,
3   96k,
3   AAC,
3   1.0,
3   s16le,
3   48000

I missed your desired output. Here it is:

$ for i in {1..3}; do
    aryvar="Audio${i}[*]"    # note the "*" not "@"
    echo "${!aryvar}"
done
0:ita, 96k, AAC, 2.0, s16le, 48000
1:rus, 128k, AAC, 2.0, s16le, 48000
2:klg, 96k, AAC, 1.0, s16le, 48000

edited Mar 11, 2016 at 21:55

answered Mar 11, 2016 at 20:54

glenn jackman

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2 Comments

Bert Over a year ago

Thank you, Glenn, too. It isn’t exactly the wanted output, but for later possibly using of simple parts of my arrays I’m going to remind your solution in the back of my mind.

glenn jackman Over a year ago

Read about variable indirection in the bash manual (skip down a couple of paragraphs)

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