bash script variable array name

Question

Hello I want to ask a question that is repeated here.

I have four servers in bash script defined like in the code below. For each server, I want to maintain the ID of the process I have started on it. Just for testing, I wanted to initialize each array with 10 20 30 40. And see if I can access these elements as expected. However I cannot access the elements. Could someone tell me what exactly I am doing wrong.

#!/bin/bash
SERVER_LIST=("server1" "server2" "server3")

for server in ${SERVER_LIST[@]} ; do
    echo $server
    arrayName=$server"process"
    echo $arrayName
    set -a "$arrayName=(10 20 30 40)"
done

current_sever=${SERVER_LIST[0]}
arrayName=$current_server"process"
# The attempt below is failing.
eval "echo Server ${current_server} \${$arrayName[*]}"
echo $(eval echo \${arrayName[*]})Server server1

server1process

It gives me output as follows -

Server server1

server1process

Can someone help please. Also can you please tell me how to append new element to the array? I tried the following, but it doesn't work -

sleep 10 &
arrayName=$current_server"process"
eval "\${$arrayName[*]}+=$!"

Please see BashFAQ/005 and BashFAQ/006.

Dennis Williamson
– Dennis Williamson

2012-08-03 22:32:51 +00:00
Commented Aug 3, 2012 at 22:32 — Dennis Williamson
– Dennis Williamson, Commented Aug 3, 2012 at 22:32
possible duplicate of Bash: Use a variable as an array name

Mechanical snail
– Mechanical snail

2012-08-06 04:13:35 +00:00
Commented Aug 6, 2012 at 4:13 — Mechanical snail
– Mechanical snail, Commented Aug 6, 2012 at 4:13

Manish Jhawar · Accepted Answer · 2012-08-25 11:08:04Z

1

Try replacing line 8:

set -a "$arrayName=(10 20 30 40)"

with:

eval "$arrayName=(10 20 30 40)"

answered Aug 25, 2012 at 11:08

Manish Jhawar

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