asked on

AIX time_last_login script

I need some help in putting a quick script to display Last Login Information of all the users on a AIX server. The format requested is something like below.

----------------------------------------------
User Gecos Last Loggedin
----------------------------------------------

User field : User ID
Gecos : User Description
Last Logged in : Readable format

On AIX, I can pull in User field and Gecos from - awk -F: '{print $1 $5}' /etc/passwd

and Last logged in from `perl -e 'print scalar localtime(1210762918);'`

How should I put everything together, so that I get an output like

----------------------------------------------
User Gecos Last Loggedin
----------------------------------------------

Thanks,

-

omarfarid

try this

echo -----------------------------------
usr=`awk -F: '{print $1 $5}' /etc/passwd`
llogin=`perl -e 'print scalar localtime(1210762918);'`
echo "$usr $llogin"
echo -----------------------------------

Ramu Shetty

ASKER

Thanks you....Omar, On the similar grounds, I wrote a script but this is quite not working...

So I have all the users on the system whose UID are starting with "p",

for i in `awk -F: '{print $1}' /etc/passwd | grep ^p`
do
x=$i
y=`lsuser -a time_last_login $i | awk -F'=' '{print $2}'`
z=`perl -e 'print scalar localtime($y);'`

a=`cat /etc/passwd |awk '{print $5}'`
echo "$x $a $z"
done

So I am trying to print out : UserID Gecod Last Login

omarfarid

Can till me what is the output of each command you run in the script?

awk -F: '{print $1}' /etc/passwd | grep ^p

lsuser -a time_last_login $i | awk -F'=' '{print $2}'

perl -e 'print scalar localtime($y);'

cat /etc/passwd |awk '{print $5}'

Ramu Shetty

ASKER

Here is the info

awk -F: '{print $1}' /etc/passwd | grep ^p
p876km
p515kr
p112qp

lsuser -a time_last_login $i | awk -F'=' '{print $2}'
1412133895
1411394010
1412109891

perl -e 'print scalar localtime($y);'
Wed Dec 31 18:00:00 1969Wed Dec 31 18:00:00 1969Wed Dec 31 18:00:00 1969

cat /etc/passwd |awk '{print $5}'
Paul Ruber
Shawn Zeiler
Mary Kom

I think the perl thing got screwed up,which is to convert the last log to the standard time format

omarfarid

when you run

echo "$x $a $z"

What do you get?

Ramu Shetty

ASKER

So when I run the Echo command I am getting some undesired output, That is the reason why I approached EE

omarfarid

what do you get?

Ramu Shetty

ASKER

Not the desired output, a blank screen scrolling with something on the bottom ...

Ramu Shetty

ASKER

Not the desired output, a blank screen scrolling with something on the bottom ...

ASKER CERTIFIED SOLUTION

omarfarid

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Ramu Shetty

ASKER

This is not working either...here is the output

p876km
Wed Dec 31 18:00:00 1969
<Shawn.M.Zeiler@XXX.com>,#9809TSMTeam@XXX.com:/home/p876km:/usr/bin/ksh
p515kr
Wed Dec 31 18:00:00 1969
<Shawn.M.Zeiler@XXX.com>,#9809TSMTeam@XXX.com:/home/p876km:/usr/bin/ksh
p112qp
Wed Dec 31 18:00:00 1969
<Shawn.M.Zeiler@XXX.com>,#9809TSMTeam@XXX.com:/home/p876km:/usr/bin/ksh

So can you please help me out on this....

Ramu Shetty

ASKER

The output I am expecting should look like...

----------------------------------------------
User Gecos Last Loggedin
----------------------------------------------

User Info can be got from -> awk -F: '{print $1}' /etc/passwd

Gecos can be got from -> awk -F: '{print $5}' /etc/passwd

Last Loggedin -> perl -e 'print scalar localtime($x);'
Where x=lsuser -a time_last_login UID | awk -F'=' '{print $2}'

Thanks...