Script to see if user has logged off
I have created this program to see if a user is logged on and it checks every minute.
if [ "$#" -ne 1 ]
then
echo "Usage: mon user"
exit 1
fi
user="$1"
until who | grep "^$user " > /dev/null
do
sleep 60
done
echo "$user has logged on"
But my question is how can I modify this program to see if a user has logged off instead of logging in?
Thanks for any help provided! Much appreciated.
bash logs users administration
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
asked Nov 13 '14 at 0:27
AphicisAphicis
62
bumped to the homepage by Community♦ 9 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
I have created this program to see if a user is logged on and it checks every minute.
if [ "$#" -ne 1 ]
then
echo "Usage: mon user"
exit 1
fi
user="$1"
until who | grep "^$user " > /dev/null
do
sleep 60
done
echo "$user has logged on"
But my question is how can I modify this program to see if a user has logged off instead of logging in?
Thanks for any help provided! Much appreciated.
bash logs users administration
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
asked Nov 13 '14 at 0:27
AphicisAphicis
62
bumped to the homepage by Community♦ 9 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
You could simply changeuntiltowhile, but I do not think this is the most efficient or elegant method to be notified of user login activity.
– jw013
Nov 13 '14 at 0:51
Couldn't just use-vwith thegrepto and print out that user is not logged in? Of course, it won't tell you whether the user had logged in and then logged out, but would just tell you whether the user is logged in at the moment.
– Sree
Nov 13 '14 at 4:52
add a comment |
I have created this program to see if a user is logged on and it checks every minute.
if [ "$#" -ne 1 ]
then
echo "Usage: mon user"
exit 1
fi
user="$1"
until who | grep "^$user " > /dev/null
do
sleep 60
done
echo "$user has logged on"
But my question is how can I modify this program to see if a user has logged off instead of logging in?
Thanks for any help provided! Much appreciated.
bash logs users administration
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
asked Nov 13 '14 at 0:27
AphicisAphicis
62
I have created this program to see if a user is logged on and it checks every minute.
if [ "$#" -ne 1 ]
then
echo "Usage: mon user"
exit 1
fi
user="$1"
until who | grep "^$user " > /dev/null
do
sleep 60
done
echo "$user has logged on"
But my question is how can I modify this program to see if a user has logged off instead of logging in?
Thanks for any help provided! Much appreciated.
bash logs users administration
bash logs users administration
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
asked Nov 13 '14 at 0:27
AphicisAphicis
62
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
asked Nov 13 '14 at 0:27
AphicisAphicis
62
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
edited Nov 13 '14 at 4:25
jasonwryan
49.9k14134188
49.9k14134188
asked Nov 13 '14 at 0:27
AphicisAphicis
62
asked Nov 13 '14 at 0:27
AphicisAphicis
62
asked Nov 13 '14 at 0:27
AphicisAphicis
62
62
bumped to the homepage by Community♦ 9 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 9 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
1
You could simply changeuntiltowhile, but I do not think this is the most efficient or elegant method to be notified of user login activity.
– jw013
Nov 13 '14 at 0:51
Couldn't just use-vwith thegrepto and print out that user is not logged in? Of course, it won't tell you whether the user had logged in and then logged out, but would just tell you whether the user is logged in at the moment.
– Sree
Nov 13 '14 at 4:52
add a comment |
1
You could simply changeuntiltowhile, but I do not think this is the most efficient or elegant method to be notified of user login activity.
– jw013
Nov 13 '14 at 0:51
Couldn't just use-vwith thegrepto and print out that user is not logged in? Of course, it won't tell you whether the user had logged in and then logged out, but would just tell you whether the user is logged in at the moment.
– Sree
Nov 13 '14 at 4:52
1
1
You could simply change
until to while, but I do not think this is the most efficient or elegant method to be notified of user login activity.– jw013
Nov 13 '14 at 0:51
You could simply change
until to while, but I do not think this is the most efficient or elegant method to be notified of user login activity.– jw013
Nov 13 '14 at 0:51
Couldn't just use
-v with the grep to and print out that user is not logged in? Of course, it won't tell you whether the user had logged in and then logged out, but would just tell you whether the user is logged in at the moment.– Sree
Nov 13 '14 at 4:52
Couldn't just use
-v with the grep to and print out that user is not logged in? Of course, it won't tell you whether the user had logged in and then logged out, but would just tell you whether the user is logged in at the moment.– Sree
Nov 13 '14 at 4:52
add a comment |
2 Answers
2
active
oldest
votes
You can filter the secure log file (assuming you are just checking for remote login). The secure log file tells you when a user logs in and logs out. Also check the lastlog command as well as the wtmp and utmp files. They'll contain historical data for user logins.
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
/var/log/secure may not be set up in all distros. I think systemd has thrown away /var/log in favor ofjournalctl. I like the idea of usingwhowbetter as this will include those who are logged in non-remotely.
– SailorCire
Nov 13 '14 at 15:13
add a comment |
Try this.
#!/bin/bash
[ "$#" -ne 1 ] && {
echo "Usage: $(basename $0) user"
exit 1
}
user="$1"
grep "^$user:" /etc/passwd > /dev/null || {
echo "There's no user called $user in this system."
exit 2
}
who | grep "^$user " > /dev/null && sleep 60
echo "$user has logged out"
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can filter the secure log file (assuming you are just checking for remote login). The secure log file tells you when a user logs in and logs out. Also check the lastlog command as well as the wtmp and utmp files. They'll contain historical data for user logins.
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
/var/log/secure may not be set up in all distros. I think systemd has thrown away /var/log in favor ofjournalctl. I like the idea of usingwhowbetter as this will include those who are logged in non-remotely.
– SailorCire
Nov 13 '14 at 15:13
add a comment |
You can filter the secure log file (assuming you are just checking for remote login). The secure log file tells you when a user logs in and logs out. Also check the lastlog command as well as the wtmp and utmp files. They'll contain historical data for user logins.
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
/var/log/secure may not be set up in all distros. I think systemd has thrown away /var/log in favor ofjournalctl. I like the idea of usingwhowbetter as this will include those who are logged in non-remotely.
– SailorCire
Nov 13 '14 at 15:13
add a comment |
You can filter the secure log file (assuming you are just checking for remote login). The secure log file tells you when a user logs in and logs out. Also check the lastlog command as well as the wtmp and utmp files. They'll contain historical data for user logins.
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
You can filter the secure log file (assuming you are just checking for remote login). The secure log file tells you when a user logs in and logs out. Also check the lastlog command as well as the wtmp and utmp files. They'll contain historical data for user logins.
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
answered Nov 13 '14 at 2:37
Tito ValentinTito Valentin
262
262
/var/log/secure may not be set up in all distros. I think systemd has thrown away /var/log in favor ofjournalctl. I like the idea of usingwhowbetter as this will include those who are logged in non-remotely.
– SailorCire
Nov 13 '14 at 15:13
add a comment |
/var/log/secure may not be set up in all distros. I think systemd has thrown away /var/log in favor ofjournalctl. I like the idea of usingwhowbetter as this will include those who are logged in non-remotely.
– SailorCire
Nov 13 '14 at 15:13
/var/log/secure may not be set up in all distros. I think systemd has thrown away /var/log in favor of
journalctl. I like the idea of using who w better as this will include those who are logged in non-remotely.– SailorCire
Nov 13 '14 at 15:13
/var/log/secure may not be set up in all distros. I think systemd has thrown away /var/log in favor of
journalctl. I like the idea of using who w better as this will include those who are logged in non-remotely.– SailorCire
Nov 13 '14 at 15:13
add a comment |
Try this.
#!/bin/bash
[ "$#" -ne 1 ] && {
echo "Usage: $(basename $0) user"
exit 1
}
user="$1"
grep "^$user:" /etc/passwd > /dev/null || {
echo "There's no user called $user in this system."
exit 2
}
who | grep "^$user " > /dev/null && sleep 60
echo "$user has logged out"
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
add a comment |
Try this.
#!/bin/bash
[ "$#" -ne 1 ] && {
echo "Usage: $(basename $0) user"
exit 1
}
user="$1"
grep "^$user:" /etc/passwd > /dev/null || {
echo "There's no user called $user in this system."
exit 2
}
who | grep "^$user " > /dev/null && sleep 60
echo "$user has logged out"
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
add a comment |
Try this.
#!/bin/bash
[ "$#" -ne 1 ] && {
echo "Usage: $(basename $0) user"
exit 1
}
user="$1"
grep "^$user:" /etc/passwd > /dev/null || {
echo "There's no user called $user in this system."
exit 2
}
who | grep "^$user " > /dev/null && sleep 60
echo "$user has logged out"
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
Try this.
#!/bin/bash
[ "$#" -ne 1 ] && {
echo "Usage: $(basename $0) user"
exit 1
}
user="$1"
grep "^$user:" /etc/passwd > /dev/null || {
echo "There's no user called $user in this system."
exit 2
}
who | grep "^$user " > /dev/null && sleep 60
echo "$user has logged out"
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
answered Nov 30 '16 at 14:58
TomaszTomasz
9,52652965
9,52652965
add a comment |
add a comment |
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1
You could simply change
untiltowhile, but I do not think this is the most efficient or elegant method to be notified of user login activity.– jw013
Nov 13 '14 at 0:51
Couldn't just use
-vwith thegrepto and print out that user is not logged in? Of course, it won't tell you whether the user had logged in and then logged out, but would just tell you whether the user is logged in at the moment.– Sree
Nov 13 '14 at 4:52