LinuxQuestions.org - Lastlog script help

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Lastlog script help

Hi everyone,

I need a little help here. I was told I need to hand over every login from every users of every nix-like systems.

On linux it was quite easy with the lastlog utility. The output looks like this:

Code:

Username        Port    From            Latest

tien                                      **Never logged in**

msimard                                    **Never logged in**

ydoucet          pts/1    10.5.5.2        Wed Sep 12 14:32:24 -0400 2007

nagios                                    **Never logged in**

nbouchard        pts/0    10.9.5.60        Mon Sep 10 14:09:30 -0400 2007

ftpimg                                    **Never logged in**

apache                                    **Never logged in**

But on Solaris I'm having a hard time. I've found lastx, which give me every last login and it looks like this:

Code:

bash-2.05# lastx -u

Username            Logins  Terminal  Host                Last Logged in 

etudia10                  3  pts/8      10.9.2.114          Thu Nov 30 16:53:55 2006

etudia08                  3  pts/16    10.9.2.102          Mon Dec 18 14:46:49 2006

scportca                  4  pts/3      10.9.18.55          Wed Sep 12 11:32:24 2007

etudia09                  3  pts/14    10.9.2.104          Thu Nov 30 16:53:29 2006

etudia11                  1  pts/6      10.9.1.71            Thu Nov 30 16:54:17 2006

scolbory                  4  pts/3      10.9.1.60            Wed Jun  7 15:23:51 2006

So far so good, except I don't get those that never logged in. Can someone help me with that? I'm no script guru. I tried something but it doesn't work:

Code:

cat /etc/passwd|cut -d ":" -f1

gives me the login name of all users on the system,

Code:

lastx -u|cut -d " " -f1

gives me the login name of all users that has ever logged in the system.

I tried

Code:

grep -v `lastx -u|cut -d " " -f1` /etc/passwd|cut -d ":" -f1

with no success.

Anyone?

Thank you very much.

I think a for loop could do you some good. Because `lastx -u|cut -d " " -f1` outputs a list so you need to take one item per time. Maybe something like:
for i in `cat /etc/passwd|cut -d ":" -f1`; do grep $i ...

Maybe this one:

Code:

for i in `cat /etc/passwd|cut -d ":" -f1`;do if (lastx -u|cut -d " " -f1|grep -q $i);then echo $i yes; else echo $i no; fi;echo bla; done

On my system it looks like:

Code:

for i in `cat /etc/passwd|cut -d ":" -f1`;do if (last|cut -d " " -f1|grep -q $i);then echo $i yes; else echo $i no; fi;echo bla; done

root yes

bla

bin no

bla

daemon no

bla

So the yes is the users that have logged in.
A littlebit cleaner to only output a list of non-logged in users:
Note that we invert the if statement here with !
This gives us a list of users that have not logged in:

Code:

for i in `cat /etc/passwd|cut -d ":" -f1`;do if ! (last|cut -d " " -f1|grep -q $i);then echo $i; fi; done

Your version:

Code:

for i in `cat /etc/passwd|cut -d ":" -f1`;do if ! (lastx -u|cut -d " " -f1|grep -q $i);then echo $i; fi; done

Since this pertains Solaris I'll move the thread there.

hey thanks muha, I'll try that first thing in the morning.

@muha, its better to break down the code for OP to understand and not cram everything into one line.
@OP. Just an example, adapt to your needs

Code:

last > outlast 

nawk 'BEGIN{FS=":"}

    FNR==NR{user[$1]; next}

    { FS=" " 

      if ( $1 in user ) print $0  

    }

' "/etc/passwd" "outlast"

thanks to all,

muha, your last non logged version is what I was looking for, thank you.

Really appreciate.

Yanik

@ghostdog74: yanik started :D

@yanik: no problem, cya :)

This is a test. Feel free to ignore.