search pattern issue

[Hakki Aydin Ucar]

The following code clause does not find what it was supposed to be; pattern returns everything in file /tmp/UserLogin. What am I missing , any advice ?

@users=`cat /etc/passwd|awk -F: '{print \$1}'`;
my $K;
for ( $K = 1; $K <= @users; $K++ )
{
open(IN,"/tmp/UserLogin");
open(OUT,">>/tmp/LogonOutput");
while ($line=) {
$pattern = "$users[$K-1]" ;
if ( $line =~ m{$pattern} ) {
print "$pattern equal to $line\n";
print OUT "$line\n";
}
}
}


Besides,
I tried m/\b$pattern/ also ,but it did not work at all.

[Horia Chirculescu]

Hello,

Why m/ ? (Multiple lines?) you only read one line at at a time, so should use s/

Best regards
Horia.

[Horia Chirculescu]

Seems that default is single line... Sorry.

[Hakki Aydin Ucar]

Horia,

m/ stand for match , but seems useless somehow here, I tried without m like this:
if ( $line =~ /\b$line/ )

it worked now.

thanks

[Horia Chirculescu]

Ohhh.. One mistake:

Move

open(IN,"/tmp/UserLogin");
open(OUT,">>/tmp/LogonOutput");

outside for ! You actually read the same data a lot of times...

Horai

[Horia Chirculescu]

And... When done, you should close the files

close(IN)
close(OUT)

Best regards,
Horia.

[Horia Chirculescu]

$line =~ /\b$line/ and it worked?!

Meaning $line was searched for $line?

[Hakki Aydin Ucar]

>Horia:$line =~ /\b$line/ and it worked?!

Meaning $line was searched for $line?

Right, it is mistake, I just noticed that put OPEN File handler out of for loop, but I am still unable to pick from file what I wanted.

[Horia Chirculescu]

Try

($line =~ m/$pattern{0}/ )

Horia.

[Horia Chirculescu]

@users=`cat /etc/passwd|awk -F: '{print \$1}'`;
my $K;
#open(IN,"/tmp/UserLogin");
open(OUT,">>/tmp/LogonOutput");

for ( $K = 1; $K <= @users; $K++ )
{
open(IN,"/tmp/UserLogin");

while ($line=)
{
$pattern = "$users[$K-1]" ;

if ( $line =~ m/$pattern{0}/ ) {
print "$pattern equal to $line\n";
print OUT "$line\n";
}
}

close(IN);

}

[Hakki Aydin Ucar]

>Horia: ($line =~ m/$pattern{0}/ )

only the first pattern working, others not working, problem is why perl picks up all pattern they are not even on the $pattern values ?

Regards,

[Hakki Aydin Ucar]

Sorry for points assigned in reverse posts but all belongs to you eventually.
Thanks, the problem solved apparently now, but another problem ,if patterns don t match , still written into OUTPUT file ??

is it weird ? something is missing in the loop ?

[Horia Chirculescu]

There should be nothing in the output file if nothing to match...

What exactly do you get in the output file in this case?

Horia.

[Horia Chirculescu]

.. and do not forget to close (OUT) when exit for (not pasted in my example code)

[Horia Chirculescu]

#!/usr/bin/perl
@users=`cat /etc/passwd|awk -F: '{print \$1}'`;
my $K;
open(OUT,">>/tmp/LogonOutput");

for ( $K = 1; $K <= @users; $K++ )
{
open(IN,"/tmp/UserLogin");
while ($line=)
{
$pattern = "$users[$K-1]" ;


if ( $line =~ m/$pattern{0}/ ) {
print "$pattern equal to $line\n";
print OUT "$line\n";
}
}
close(IN);

}

close(OUT);

[Hakki Aydin Ucar]

>Horia:There should be nothing in the output file if nothing to match...
What exactly do you get in the output file in this case?

this is from output file:
LOGIN Sep 17 2010 18:03:09
root Sep 17 2010 18:03:24
student Sep 20 2010 13:55:45


LOGIN and root are not in the @users list but goes to output file ??

[Hakki Aydin Ucar]

Horia,

I tried to change environment and seems its working now , so my test server has some problems by Perl sources ??
I will notify the solution after completed tests.
Thanks
Hakki

[Horia Chirculescu · Accepted solution]

I believe user "root" should be in /etc/passwd so will be present also in @users list.

This could be true also for the "student" user.

Looks strange the presence of LOGIN.

grep LOGIN /etc/passwd?


Can you attach the file /tmp/UserLogin ?

Horia.

[Horia Chirculescu]

>I tried to change environment

What exactly did you changed in the env. ?

Horia.

[Hakki Aydin Ucar]

>Horia:I believe user "root" should be in /etc/passwd so will be present also in @users list.

No, I took it out from passwd file with "grep -v"

I meant another server with environment change.

Last script is here (working great now!)
@users=`cat /etc/passwd|awk -F: '{print \$1}'|grep -vE 'root|nwauto|daemon|bin|sys|adm|uucp|lp|www|webadmi
n|hpdb|informix|nuucp|smbnull|mysql|nwcron|goglobal|alop|trop|dbadmin|c7op|sysadmin`;
my $K;
open(OUT,">>/tmp/LogonOutput");
for ( $K = 1; $K <= @users; $K++ )
{
open(IN,"/tmp/UserLogin");
while ($line=)
{
$pattern = "$users[$K-1]" ;
if ( $line =~ m/\b$pattern{0}/ ) {
print "$pattern equal to $line\n";
print OUT "$line";
}
}
close(IN);
}
close(OUT);

[Hakki Aydin Ucar]

>Horial: if ( $line =~ m/$pattern{0}/ )

what exactly means that {0} in $pattern{0} ?

[Horia Chirculescu]

Read about quantifiers in (and search google for more):

http://www.perlmonks.org/?node_id=967

Best regards,
Horia.

[James R. Ferguson]

Hi Hakki:

There are a number of things wrong in your script.

First, you count starting an an index of one. Perl is zero relative. Your 'for' loop thus skips the first entry in the '/etc/passwd' file and attempts to process a non-existent one when it compares '$K<='.

Contrast your code to this:

#!/usr/bin/perl
use strict;
use warnings;
my ( @users, $K );
@users=`cat /etc/passwd|awk -F: '{print \$1}'`;
for ( $K = 0; $K < @users; $K++ ) {
print $users[$K];
}

...

Next, the above is horribly un-Perlish. You start a external process to fill an array and then immediately fork a 'cat' and an 'awk' when the 'awk' would do alone (if this were a shell program). There is *no* need for any of this in a Perl script.

Instead do something like:

#!/usr/bin/perl
use strict;
use warnings;
my $fh;
my $file = '/etc/passwd';
my @users;
open( $fh, '<', $file ) or die "Can't open '$file': $!\n";
while (<$fh>) {
push @users, (split /:/) [0];
}
for my $user (@users) {
print "$user\n";
}

I would urge you to use the three-argument form of 'open()' too.

Regards!

...JRF...

[Hakki Aydin Ucar]

>JRF:

Great and very Perlish solution :) Thanks.
What method do you recommend to get rid of unwanted system users like grep -v is for All lines but those matching are printed ?

All lines but those matching are printed.

@users=`cat /etc/passwd|awk -F: '{print \$1}'|grep -vE 'root|nwauto|daemon|bin|sys|adm|uucp|lp|www|we
badmin|hpdb|informix|nuucp|smbnull|mysql|nwcron|goglobal|alop|trop|dbadmin|c7op|sysadmin|ftp`;

Best Regards,
Hakki

[James R. Ferguson]

Hi (again) Hakki:

> What method do you recommend to get rid of unwanted system users

One way is to create a hash of exclusions. Then, skip the processing of any user that exists in the hash:

# cat ./dousers
#!/usr/bin/perl
use strict;
use warnings;
my $fh;
my $file = '/etc/passwd';
my @users;
my %exclude;
@exclude{qw( root daemon bin sys adm uucp lp nuucp hpdb nobody www)} = ();
open( $fh, '<', $file ) or die "Can't open '$file': $!\n";
while (<$fh>) {
push @users, ( split /:/ )[0];
}
for my $user (@users) {
print "$user\n" unless exists $exclude{$user};
}
1;

Regards!

...JRF...