Bash: dynamically assigning array to a new variable

Question

I have two arrays which I would like to dynamically assign to a variable depending on user input

ARRAY_ONE=('one' 'two')
ARRAY_TWO=('three' 'four')

Suppose that $opt can be either ONE or TWO depending on user input. I have a variable ARRAY_THREE that I would like to assign to the content of either ARRAY_ONE or ARRAY_TWO depending on the value of $opt.

The following snippet does not work, as it only takes the element in the first position of the assigned array:

TEMP=ARRAY_$opt
ARRAY_THREE=${!TEMP}
echo $ARRAY_THREE     # 'one'

anubhava · Accepted Answer · 2013-06-18 08:41:58Z

4

Change your code to:

TEMP=ARRAY_$opt[@]
ARRAY_THREE=(${!TEMP})
echo ${ARRAY_THREE[@]}

OUTPUT:

three four

EDIT:

Live Demo: http://ideone.com/hocG24

edited Jun 18, 2013 at 8:41

answered Jun 18, 2013 at 8:35

anubhava

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2 Comments

anubhava Over a year ago

@perreal: Pls check the linked demo.

anubhava Over a year ago

@perreal: Edited the demo to add echo ${ARRAY_THREE[0]} and echo ${ARRAY_THREE[1]}

Adrian Frühwirth · Accepted Answer · 2013-06-18 08:30:36Z

3

$ opt=ONE
$ TEMP="ARRAY_${opt}[@]"
$ ARRAY_THREE=( "${!TEMP}" )
$ set|grep ^ARRAY_
ARRAY_ONE=([0]="one" [1]="two")
ARRAY_THREE=([0]="one" [1]="two")
ARRAY_TWO=([0]="three" [1]="four")

answered Jun 18, 2013 at 8:30

Adrian Frühwirth

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Comments

ingyhere · Accepted Answer · 2020-09-03 08:22:51Z

0

Here is my solution that does not use a temp variable:

#!/bin/bash

foo_1=(fff ddd) ;
foo_2=(ggg ccc) ;

for i in 1 2 ;
do
    eval mine=( \${foo_$i[@]} ) ;
    echo ${mine[@]} ;
done ;

edited Sep 3, 2020 at 8:22

answered Jan 5, 2017 at 20:41

ingyhere

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Bash: dynamically assigning array to a new variable

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OUTPUT:

Live Demo: http://ideone.com/hocG24

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3 Answers 3

OUTPUT:

Live Demo: http://ideone.com/hocG24

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