small shell script which users have went in and out

Question

I needed to write a small script Shell bash script

The task is as follows:

"Develop a program that outputs information about users in the system after a certain time interval: who went in, who went out."

I tried to do this:

#!/bin/bash
while [ true ]; do
    clear
    date
    who
    sleep 5
done

then i tried like this:

#!/bin/bash
while [ true ]; do
    clear
    date
    $USER
    sleep 5
done

But everything is still the same.Only the date is displayed.(I run with windows through git bash)

What am I doing wrong?

As far as I understand, the commands that I have prescribed should simply show those who are in the system. But how exactly to deduce who entered and who left?

I thank you in advance for your help, I'm quite new to this and don't understand much, sorry.

Answer 1

You could just follow the file where logins are logged and filter the output, e.g. like this:

tail -f /var/log/auth.log | egrep "(Accepted publickey|Accepted password|Disconnected from user)"

on a debian system this will catch all lines containing ssh-logins either with password or public key, you will have to change the path/name of the logfile (/var/log/auth.log) and the match-phrases to your needs.

the output would then be something like:

Sep  6 11:43:28 myserver sshd[19052]: Accepted password for myuser from 85.237.30.84 port 43890 ssh2
Sep  6 11:43:29 myserver sshd[19522]: Disconnected from user myuser 85.237.30.84 port 43890
Sep  6 11:43:51 myserver sshd[20287]: Accepted publickey for myuser from 85.237.30.84 port 38334 ssh2: ED25519 SHA256:/****
Sep  6 11:43:57 myserver sshd[20420]: Disconnected from user myuser 85.237.30.84 port 38334

the -f switch on tail lets you follow the file until you hit <ctrl>-c you may also list all matching lines in the logfile by replacing tail -f with cat which reads the whole file.