iteration of for loop in shell script

Question

I have a simple script that uses a for loop where I want to take the value of a variable and assign it to another command and kill the screen. The script goes as

#!/usr/bin/sh/
echo 'I am going to kill the screens now';
for j in $(seq 1 1 10)
do
    i=$(grep -oP '.*?(?=\.)');
    echo $i
    screen -S $i -p 0 -X quit 
done <file1.txt
echo 'exiting...'

file1.txt

There are screens on:

    4974.eth2   (Detached)
    5105.eth11  (Detached)
    4990.eth3   (Detached)
    5006.eth5   (Detached)
    5054.eth8   (Detached)
    5070.eth9   (Detached)
    5038.eth7   (Detached)
    5022.eth6   (Detached)
8 Sockets in /var/run/screen/S-root.

So here, I am iterating through 10 lines and assigning the value of i=whatever before a period(.). eg. 4974 . Then placing 4974 in screen command and killing it. The problem here is, it is reading and grepping all the lines at a time and in the value of i it places as 4974 5105 4990 5006 5054 5070 5038 5022 in the screen command and it obviously it cannot find a screen id like that and doesn't kill any screen. How do i iterate the loop such that at a time it takes one id ie.value of i=4974 and place only that in the screen command so that it will kill it.. Let me know if you understand my question correctly.

Answer 1

Try this:

awk -F'[. ]+' '/Detached/ {print $2}' file.txt | while read i
do
    echo "$i"
    screen -S "$i" -p 0 -X quit 
done

Answer 2

Instead of iterating over a counter, iterate over your results directly:

set -f # disable globbing, so globs in grep's output aren't expanded
echo 'I am going to kill the screens now' >&2
for i in $(grep -oP '.*?(?=\.)'); do
    echo "$i"
    screen -S "$i" -p 0 -X quit 
done <file1.txt
echo 'exiting...' >&2