delete multiple users
I am the root user and I am setting up a menu for another user to use. This other user will only get this menu.
There are two options that are interlinked: the first option is to search users. The code I got is:
last | awk '{print $1,$4,$5,$6,$7} '
I have checked this code and it works, it shows me the usernames and the day they last logged on.
For the second option: I want to be able to set a date, and them delete users who haven't been active since that date, using the output of the above command.
I am using Linux Mint and Vim text editor.
awk last
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
asked Dec 5 '14 at 12:26
user2995836user2995836
484
bumped to the homepage by Community♦ 23 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 am the root user and I am setting up a menu for another user to use. This other user will only get this menu.
There are two options that are interlinked: the first option is to search users. The code I got is:
last | awk '{print $1,$4,$5,$6,$7} '
I have checked this code and it works, it shows me the usernames and the day they last logged on.
For the second option: I want to be able to set a date, and them delete users who haven't been active since that date, using the output of the above command.
I am using Linux Mint and Vim text editor.
awk last
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
asked Dec 5 '14 at 12:26
user2995836user2995836
484
bumped to the homepage by Community♦ 23 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
A question—i.e. a sentence with a question mark (?)—would greatly help to indicate what you try to ask.
– Anthon
Dec 5 '14 at 13:41
i want to know a way to delete multiple users who haven't logged in for a long period of time
– user2995836
Dec 5 '14 at 13:56
correction i want to enter a date then delete users who have been inactive since that date
– user2995836
Dec 5 '14 at 14:03
add a comment |
I am the root user and I am setting up a menu for another user to use. This other user will only get this menu.
There are two options that are interlinked: the first option is to search users. The code I got is:
last | awk '{print $1,$4,$5,$6,$7} '
I have checked this code and it works, it shows me the usernames and the day they last logged on.
For the second option: I want to be able to set a date, and them delete users who haven't been active since that date, using the output of the above command.
I am using Linux Mint and Vim text editor.
awk last
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
asked Dec 5 '14 at 12:26
user2995836user2995836
484
I am the root user and I am setting up a menu for another user to use. This other user will only get this menu.
There are two options that are interlinked: the first option is to search users. The code I got is:
last | awk '{print $1,$4,$5,$6,$7} '
I have checked this code and it works, it shows me the usernames and the day they last logged on.
For the second option: I want to be able to set a date, and them delete users who haven't been active since that date, using the output of the above command.
I am using Linux Mint and Vim text editor.
awk last
awk last
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
asked Dec 5 '14 at 12:26
user2995836user2995836
484
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
asked Dec 5 '14 at 12:26
user2995836user2995836
484
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
edited Dec 5 '14 at 13:42
Alaa Ali
9602920
9602920
asked Dec 5 '14 at 12:26
user2995836user2995836
484
asked Dec 5 '14 at 12:26
user2995836user2995836
484
asked Dec 5 '14 at 12:26
user2995836user2995836
484
484
bumped to the homepage by Community♦ 23 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♦ 23 mins ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
A question—i.e. a sentence with a question mark (?)—would greatly help to indicate what you try to ask.
– Anthon
Dec 5 '14 at 13:41
i want to know a way to delete multiple users who haven't logged in for a long period of time
– user2995836
Dec 5 '14 at 13:56
correction i want to enter a date then delete users who have been inactive since that date
– user2995836
Dec 5 '14 at 14:03
add a comment |
A question—i.e. a sentence with a question mark (?)—would greatly help to indicate what you try to ask.
– Anthon
Dec 5 '14 at 13:41
i want to know a way to delete multiple users who haven't logged in for a long period of time
– user2995836
Dec 5 '14 at 13:56
correction i want to enter a date then delete users who have been inactive since that date
– user2995836
Dec 5 '14 at 14:03
A question—i.e. a sentence with a question mark (
?)—would greatly help to indicate what you try to ask.– Anthon
Dec 5 '14 at 13:41
A question—i.e. a sentence with a question mark (
?)—would greatly help to indicate what you try to ask.– Anthon
Dec 5 '14 at 13:41
i want to know a way to delete multiple users who haven't logged in for a long period of time
– user2995836
Dec 5 '14 at 13:56
i want to know a way to delete multiple users who haven't logged in for a long period of time
– user2995836
Dec 5 '14 at 13:56
correction i want to enter a date then delete users who have been inactive since that date
– user2995836
Dec 5 '14 at 14:03
correction i want to enter a date then delete users who have been inactive since that date
– user2995836
Dec 5 '14 at 14:03
add a comment |
2 Answers
2
active
oldest
votes
You can do it like so:
root@host# lastlog -b Num_Days_Since_Last_Login | egrep -v "^Username|Never logged in" | awk '{print $1}' | xargs -i userdel {}
Where Num_Days_Since_Last_Login is an integer number of days since last login...
answered Dec 5 '14 at 15:15
KirillKirill
323
add a comment |
A complete solution :
#!/bin/bash
maxdate=$(date -d "$1" +%s) || exit 1
daysdate=$((($(date +%s)-maxdate)/(3600*24)))
LANG=C lastlog -b $daysdate -u 1000- |
awk 'NR>1{print $1}' |
xargs -n1 echo userdel
Test it and remove echo to do it forReal™.
Example usage :
./script.sh 'Fri Dec 5 17:00:06 CET 2013'
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
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 do it like so:
root@host# lastlog -b Num_Days_Since_Last_Login | egrep -v "^Username|Never logged in" | awk '{print $1}' | xargs -i userdel {}
Where Num_Days_Since_Last_Login is an integer number of days since last login...
answered Dec 5 '14 at 15:15
KirillKirill
323
add a comment |
You can do it like so:
root@host# lastlog -b Num_Days_Since_Last_Login | egrep -v "^Username|Never logged in" | awk '{print $1}' | xargs -i userdel {}
Where Num_Days_Since_Last_Login is an integer number of days since last login...
answered Dec 5 '14 at 15:15
KirillKirill
323
add a comment |
You can do it like so:
root@host# lastlog -b Num_Days_Since_Last_Login | egrep -v "^Username|Never logged in" | awk '{print $1}' | xargs -i userdel {}
Where Num_Days_Since_Last_Login is an integer number of days since last login...
answered Dec 5 '14 at 15:15
KirillKirill
323
You can do it like so:
root@host# lastlog -b Num_Days_Since_Last_Login | egrep -v "^Username|Never logged in" | awk '{print $1}' | xargs -i userdel {}
Where Num_Days_Since_Last_Login is an integer number of days since last login...
answered Dec 5 '14 at 15:15
KirillKirill
323
answered Dec 5 '14 at 15:15
KirillKirill
323
answered Dec 5 '14 at 15:15
KirillKirill
323
answered Dec 5 '14 at 15:15
KirillKirill
323
323
add a comment |
add a comment |
A complete solution :
#!/bin/bash
maxdate=$(date -d "$1" +%s) || exit 1
daysdate=$((($(date +%s)-maxdate)/(3600*24)))
LANG=C lastlog -b $daysdate -u 1000- |
awk 'NR>1{print $1}' |
xargs -n1 echo userdel
Test it and remove echo to do it forReal™.
Example usage :
./script.sh 'Fri Dec 5 17:00:06 CET 2013'
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
add a comment |
A complete solution :
#!/bin/bash
maxdate=$(date -d "$1" +%s) || exit 1
daysdate=$((($(date +%s)-maxdate)/(3600*24)))
LANG=C lastlog -b $daysdate -u 1000- |
awk 'NR>1{print $1}' |
xargs -n1 echo userdel
Test it and remove echo to do it forReal™.
Example usage :
./script.sh 'Fri Dec 5 17:00:06 CET 2013'
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
add a comment |
A complete solution :
#!/bin/bash
maxdate=$(date -d "$1" +%s) || exit 1
daysdate=$((($(date +%s)-maxdate)/(3600*24)))
LANG=C lastlog -b $daysdate -u 1000- |
awk 'NR>1{print $1}' |
xargs -n1 echo userdel
Test it and remove echo to do it forReal™.
Example usage :
./script.sh 'Fri Dec 5 17:00:06 CET 2013'
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
A complete solution :
#!/bin/bash
maxdate=$(date -d "$1" +%s) || exit 1
daysdate=$((($(date +%s)-maxdate)/(3600*24)))
LANG=C lastlog -b $daysdate -u 1000- |
awk 'NR>1{print $1}' |
xargs -n1 echo userdel
Test it and remove echo to do it forReal™.
Example usage :
./script.sh 'Fri Dec 5 17:00:06 CET 2013'
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
edited Dec 5 '14 at 16:40
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
answered Dec 5 '14 at 16:21
Gilles QuenotGilles Quenot
16k13951
16k13951
add a comment |
add a comment |
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A question—i.e. a sentence with a question mark (
?)—would greatly help to indicate what you try to ask.– Anthon
Dec 5 '14 at 13:41
i want to know a way to delete multiple users who haven't logged in for a long period of time
– user2995836
Dec 5 '14 at 13:56
correction i want to enter a date then delete users who have been inactive since that date
– user2995836
Dec 5 '14 at 14:03