Concatenate arrays using iteration in name with bash

Question

I would like to concatenate unlimited numbers of arrays using shortest lines possible, so for this I did the code below:

#!/bin/bash
declare -a list1=("element1")
declare -a list2=("element2")
declare -a list3=("element3")
declare -a list4=("element4")
declare -a list
for i in {1..4}
do
   list=( ${list[@]} ${list$i[@]} )
done
echo ${list[*]}

But the code above is not working because $i is not seen as variable and the error is: ${list$i[@]} bad substitution

using shortest lines then why do you use so many spaces? to concatenate unlimited numbers you want to output them or store them in an array? Please be specific. — KamilCuk
– KamilCuk, Commented Jan 18, 2022 at 9:23
Unlimited? Do you really want to create an unlimited number of variables? — choroba
– choroba, Commented Jan 18, 2022 at 9:27
do not care of spaces, I used for better view. I want to have all elements in one array first, surely I will use elements later but how could I use a variable as number in array name for this concatenation? — antonio1
– antonio1, Commented Jan 18, 2022 at 9:27
just saying unlimited number of arrays, this is limited to 10 at first but later I want to add mores so they could be 100 — antonio1
– antonio1, Commented Jan 18, 2022 at 9:28

choroba · Accepted Answer · 2022-01-18 16:32:20Z

1

You can use variable indirection:

for i in {1..4} ; do
    ref="list$i[@]"
    list+=("${!ref}")
done
echo "${list[@]}"

edited Jan 18, 2022 at 16:32

answered Jan 18, 2022 at 9:29

choroba

245k27 gold badges221 silver badges304 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Gordon Davisson Over a year ago

You need double-quotes around ${!ref} -- without them, it'll be subject to the usual word-splitting and wildcard-expansion messiness.

antonio1 Over a year ago

not working, I have no error but when printing using echo I have just first element1

choroba Over a year ago

@antonio1: Weird, it works for me with your example.

KamilCuk · Accepted Answer · 2022-01-18 09:40:31Z

0

The following code outputs all 4 lists concatenated together.

eval echo \${list{1..4}[*]}

This code runs filename expansion over the result of list elements (* is replaced by filenames). Consider sacrificing 4 characters and doing \"\${list{1..4}[*]}\".

Note that eval is evil https://mywiki.wooledge.org/BashFAQ/048 and such code is confusing. I wouldn't write such code in a real script - I would definitely use a loop. Use shellcheck to check your scripts.

edited Jan 18, 2022 at 9:40

answered Jan 18, 2022 at 9:27

KamilCuk

146k8 gold badges84 silver badges154 bronze badges

2 Comments

antonio1 Over a year ago

I'm sorry but something is not good, finally I get all elements in one array but they don't have double quotas anymore and when I try to use the final array with a for loop to get elements one by one it is seen all elements as one. This is not good.

KamilCuk Over a year ago

The code I presented does not put elements in an array. It outputs them. As I understood, the goal was to output them in shortest lines possible. finally I get all elements in one array then you are not using the code I presented in this answer (?). Do you actually need for it be shortest as possible? See the other answer - it properly handles quoting. Check your scripts with shellcheck

Collectives™ on Stack Overflow

Concatenate arrays using iteration in name with bash

2 Answers 2

3 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related