How and where does bash remove an array element?

Question

linux

I decided to make a script that will watch the output of the bot upon completion of it and look for "reboot" there in order to restart it and start all over again. I have not yet figured out how the script itself will be restarted from it, but not about that now (although if you suggest how to implement this, I will only be grateful). So far, I'm testing on this js script:

console.log("test");
console.log("test1");

The bash script itself:

read -p "Enter the programm's path with .js: " path
    cd $path
    echo "node $path/main.js"
    nodeLogs="$(node $path/main.js)"
        echo $nodeLogs
        IFS="\n" read -a nodeLogsArr <<< "$nodeLogs"
            echo "${nodeLogsArr[1]}"
            for element in "${nodeLogsArr[@]}"; do
                echo "$element"
            done
            for logs in ${nodeLogsArr[@]}; do
                if (($logs="reboot")); then
                    echo "U couldn't see it!!!"
                    #git pull
                else
                    echo "$logs"
                fi
            done

It turns out that the script, after translating the multiline string into array elements, puts the second element "test1" somewhere. I studied partitioning using IFS according to this guide . I did it, it seems, as it should, but it does not want to work ...

Script output:

$ ./test
Enter the programm's path with .js: .
node ./main.js
test test1

test
test

Maybe I'm trying to split the string incorrectly?

Thanks in advance for your reply

Answer 1

in a simple case:

$ x='test1 test2'
$ y=($x)
$ echo ${y[0]} И ${y[1]}
test1 И test2

if, for example, we have a newline separator, then:

$ x='test 1
test 2'
$ q="$IFS";IFS=$'\n';y=($x);IFS="$q"
$ echo ${y[0]} И ${y[1]}
test 1 И test 2