If you want to receive a notification over XMPP everytime someone logs in to your machine, follow these steps:

  1. Install sendxmpp and libpam-script
  2. Create /usr/local/sbin/pam_script_ses_open with the following content:
    #! /bin/sh
    exit 0
  3. Execute chmod 755 /usr/local/sbin/pam_script_ses_open as pam-script checks the permissions to ensure all processes can execute this script. We will restrict the access on the following file as the sshd process has root privileges.
  4. Create /usr/local/sbin/notify-login-over-xmpp.sh with the following content:
    #! /bin/sh
    echo "Login of user \"${PAM_USER}@$(hostname)\" from ${PAM_RHOST} over $PAM_SERVICE tty $PAM_TTY" | sendxmpp -t <Diese E-Mail-Adresse ist vor Spambots geschützt! Zur Anzeige muss JavaScript eingeschaltet sein.> --tls-ca-path=/etc/ssl/certs/ -u <userlogin> -j <server> -p <password> -f /dev/null
  5. Execute chmod 700 /usr/local/sbin/notify-login-over-xmpp.sh to protect your password
  6. If you use apparmor,  add the following to /etc/apparmor.d/usr.sbin.sshd to enable the sshd to execute these scripts:
    /usr/local/sbin/pam_script_ses_open rUx,
    /usr/local/sbin/notify-login-over-xmpp.sh rUx,
  7. Add the following at the end of /etc/pam.d/sshd:
    session    required     pam_script.so dir=/usr/local/sbin/ onerr=success
  8. Check if you can still login without closing the current session as misconfiguring PAM can lock you out of your machine.