topic Re: Test Processing in Operating System - HP-UX

Test Processing

Waqar Razi — Tue, 26 Aug 2008 19:02:09 GMT

I have 3 passwd files. Some of the users exist in 3 files (they are from 3 different servers) and some of them only exist in only one file.

I want to make a single file taking users entries from 3 files combine into a single file. For users who exist in all the 3 servers, I want to list their 3 uids infront of their user names. Can some one give me some direction how to do that?

Re: Test Processing

Steven E. Protter — Tue, 26 Aug 2008 20:08:49 GMT

Shalom,

cat passwd | awk -F: '{print $1}'

Use this command to make lists.

diff the files and use sort -u to come up with a single, valid master user list.

Then combine the files. Probably best to cat them into one file and delete the dups.

SEP

Re: Test Processing

Prashanth Waugh — Wed, 27 Aug 2008 04:00:14 GMT

Hi Razi,

U can use the below command for 3 passwd file. then take the output of each file into same xl sheet then sort it by name. After that u can make one file .But before doing any activity like replacing the file first take the backup of that file.

the command is
cat /etc/passwd | awk -F: '{print $1 ,$3}'

Best Luck
Atul

Re: Test Processing

Rasheed Tamton — Wed, 27 Aug 2008 08:33:09 GMT

Is it what you want:

awk -F: '{print $3, $1}' firstfile secfile thirdfile |sort -k2 > userlist

Re: Test Processing

Waqar Razi — Wed, 27 Aug 2008 12:31:40 GMT

Let me give you more detail:

Lets say I have a user called steve on three servers having different uids like 108, 120, 139 on three different servers but same username steve on all of them, and on the other hand, lets say I have another user kami on just one server having uid 159, I want my resulting file to look like that.

steve 108 120 130
kami 159

Can some one help me in this?

Re: Test Processing

James R. Ferguson — Wed, 27 Aug 2008 12:47:49 GMT

Hi Wager:

# cat .allusers
#!/usr/bin/perl
use strict;
use warnings;
my ( %users, $name, $uid );
while (<>) {
my ( $name, $uid ) = ( split /:/ )[ 0, 2 ];
push( @{ $users{$name} }, $uid );
}
for $name ( sort keys %users ) {
print "$name: @{$users{$name}}\n";
}
1;

...run, passing any number of files as arguments, like:

# ./allusers /etc/passwd /etc/passwd.server2 /etc/passwd.server3

...output looks like:

ablebody: 300
somebody: 333
wagar: 111 1221
zebra: 101 102 101

Regards!

...JRF...

Re: Test Processing

Tim Nelson — Wed, 27 Aug 2008 12:52:17 GMT

A basic way. I am sure there is something more elegant. Might just give you some ideas.

awk -F: '{print $1 " " $3}' passwd1|while read username uid
do
echo "File1: $uid $username"
echo "File2: \c"
awk -F: '$1 ~ "$username" {print $1 " " $3}' passwd2
echo "File3: \c"
awk -F: '$1 ~ "$username" {print $1 " " $3}' passwd3
done

Re: Test Processing

Dennis Handly — Thu, 28 Aug 2008 09:29:24 GMT

Try this:
sort passwd1 passwd2 passwd3 | awk -F: '
BEGIN { getline; name = $1; uid = $3 }
{
if ($1 == name) {
uid = uid " " $3 # concat
} else {
print name, uid # print last
name = $1; uid = $3
}
}
END { print name, uid } '

Re: Test Processing

Waqar Razi — Thu, 28 Aug 2008 11:38:04 GMT

Lst script is working fine, thank you very much guys for your help and support.

I need one more help, Lets say if I wanna add when that particular user logs in last time, how can I add that to this script.

Lets say if I have a user steve and he logged in for the last time one week before, output should be like that:

Steve uid1 uid2 uid3 19-AUG-08