topic Re: User last successful login in Operating System - HP-UX

User last successful login

Ngoh Chean Siung — Tue, 26 Oct 2004 02:30:27 GMT

Hi,

Is there any way or script to get all the users last successful login & pipe to a text file?

Example the data captured is:
peter|24Oct2004
andrew|16Oct2004

regards.

Re: User last successful login

Michael Schulte zur Sur — Tue, 26 Oct 2004 02:39:32 GMT

Hi,

just use > after the command you used to get the data captured.
Your question is a bit vague.
You can use cut or awk with field separator | to single out fields.

greetings,

Michael

Re: User last successful login

G. Vrijhoeven — Tue, 26 Oct 2004 02:40:50 GMT

Hi,

You could use the last command.

e.g.

for i in $(cat /etc/passwd | awk -F: '( $3 > 100 ) { print $1 } ')
do
last -1 $i | awk '{ print $1"|"$5,$4 }'
done

Regards,

Gideon

Re: User last successful login

Franky_1 — Tue, 26 Oct 2004 02:51:06 GMT

Hi,

you can use the "last" command which gives you your desired output

only last will give you the successful logins of all users

if you use last it'll give you the info only for that named user
You can customize the output using awk

eg

last |awk '{print $1,$4,$5,$6}'

lastb would be the contrary (unsuccessful logins) btw

Regards

Franky

Re: User last successful login

Fred Ruffet — Tue, 26 Oct 2004 02:51:17 GMT

As long as I think your main problem is to get all logins, you can use the "last" command as root. It will give you all logins for all users.

Parsing it with perl like this :
$ll="";
open(STDIN,"last|sort -k 1|");
while () {
chomp;
$cl=$_;
($user,@rest)=split / */,$cl;
($luser,@rest)=split / */,$ll;
if ($user eq $luser) {
$ll=$cl;
} else {
print "$ll\n";
$ll=$cl;
}
}
print "$cl\n";

will give you last line for each user. Modify it to get the output you want.

Regards,

Fred

Re: User last successful login

Ngoh Chean Siung — Tue, 26 Oct 2004 04:31:05 GMT

Hi,

What I want is to get the LATEST last login for each users. Example andrew has login to server in July and Aug, then the script will only retrieve a line of user id & the month is Aug. (No duplicated)

Any idea?

regards.

Re: User last successful login

Fred Ruffet — Tue, 26 Oct 2004 04:39:43 GMT

... That's exactly what my script did...

Regards,

Fred

Re: User last successful login

G. Vrijhoeven — Tue, 26 Oct 2004 04:42:22 GMT

Hi,

last -1 username

check my previouse post, may be it has some errors but it does a grep on user stings in the passwd file, ( only users >100 as uid) and issue a last -1 on that user. Redirection to a file .. done >>file

Regards,

Gideon

Re: User last successful login

Ngoh Chean Siung — Tue, 26 Oct 2004 05:23:29 GMT

Re: User last successful login

G. Vrijhoeven — Tue, 26 Oct 2004 05:54:09 GMT

Hi,

In order to sort, use the sort command
...
done | sort >>file

Why the only the "|" appears.. this can be the case when the user never logged in.
you can solv this by
..
done | grep -v ^| sort >>file

Regards,

Gideon

Re: User last successful login

Ngoh Chean Siung — Tue, 26 Oct 2004 06:09:45 GMT

Hi,

It prompts me this error when I added the line that you proposed

# sh user.sh
grep: illegal option -- ^
usage: grep [-E|-F] [-c|-l|-q] [-bhinsvx] -e pattern_list...[-f pattern_file...] [file...]

usage: grep [-E|-F] [-c|-l|-q] [-bhinsvx] [-e pattern_list...]-f pattern_file...[file...]

usage: grep [-E|-F] [-c|-l|-q] [-bhinsvx] pattern [file...]

Btw, is it able to show the year login as well?

regards.

Re: User last successful login

G. Vrijhoeven — Tue, 26 Oct 2004 06:17:39 GMT

Sorrie,

done | grep -v "^|" | sort >> file

As far as i know last does not print a year.

HTH,

Gideon

Re: User last successful login

Fred Ruffet — Tue, 26 Oct 2004 07:02:32 GMT

Errors on my script... maybe because it is a perl script. Not a shell one :)

Regards,

Fred

Re: User last successful login

Hein van den Heuvel — Tue, 26 Oct 2004 07:26:38 GMT

'one liner' in perl:

last | perl -e 'while(<>){($u,$x,$x,$m,$d)=split; $all{$u}="$m $d" unless ($all{$u})} foreach (sort keys %all) {print "$_|$all{$_}\n" if ($_)}'

- loop through input
- split by whitespace, remembering user, month, day
- remember date for user unless that user has a day already
- when all input processed, report on all remembered users, sorted.

no, I do not knwo how a year change looks like. But a perl script like above is readily expanded to have a rolling months window where every month in this year prints the current year, and every 'future' month prints the last year.

Hein

Re: User last successful login

Muthukumar_5 — Tue, 26 Oct 2004 08:02:00 GMT

We can collect all normal users ( except system users) with listusers so that,

for user in `listusers | awk '{ print $1 }'`
do
last -1 $user | awk '{ print $1"|"$5$4 }'
done

We can not get year informations from last output there.

HTH.

Re: User last successful login

Muthukumar_5 — Tue, 26 Oct 2004 08:07:35 GMT

To get current year and more efficient manner then use as,

# for user in `logins -u | awk '{ print $1 }'`
do
last -1 $user | grep -v '^wtmp begins' | awk '{ printf $1"|"$5$4;system("date +'%Y'") }'
done

Some user will not contain login details. That will show as wtmp begin.. information with log there.

HTH.

Re: User last successful login

Hein van den Heuvel — Tue, 26 Oct 2004 08:22:01 GMT

I needed a coffee break.. below a potential solution with year handling. It compares todays month and day ($M, $D) with the one in the last record ($m, $d) and adjust the year if deemed in the past.

test.p:
($x,$x,$x,$D,$M,$Y)=localtime;
while(<>){
($u,$x,$x,$m,$d)=split;
next if ($all{$u});
$y = $Y + 1900;
$i = index("JanFebMarAprMayJunJulAugSepOctNovDec",$m)/3;
next if ($i < 0);
$y-- if ($i > $M);
$y-- if ($i == $M && $d > $D);
$all{$u}="|$m $d $y";
}
foreach (sort keys %all) {
print "$_$all{$_}\n" if ($_);
}

Usage:

last | perl test.p

Cheers,
Hein.