simple perl and regex

phlx · 12-02-2004, 04:34 PM

Hi I am trying to code a simple script in perl
I want to call it changeshell.
What it will do, is scan through the /etc/passwd file, and search entry which contains a shell, and change it to an inputted shell.
How can I scan w/ regex and replace a certain field?

sigsegv · 12-02-2004, 05:25 PM

Code:

while(<FILEHANDLE>) {
        $line =~ s!/bin/bash\n$!/bin/false\n!;
        push(@output, $line);
}

foreach (@output) {
        print(OUT);
}

That will require fleshing out obviously.

hk_linux · 12-02-2004, 10:51 PM

You can use the command line script,

perl -pi -e s?/bin/bash\$?WATEVERINPUT? /etc/passwd

The \$ will ensure only the entries at the end of the line will be replaced
This will directly overwrite the /etc/passwd.

phlx · 12-02-2004, 11:06 PM

well i dont want to overwrite it completely just swap the entry of a said user

sigsegv · 12-02-2004, 11:26 PM

Quote:

Originally posted by hk_linux
This will directly overwrite the /etc/passwd.

Yes ... yes it will ... make sure to get it right the first time

You'll also want the correct shell syntax, which is: perl -pi -e 's#/bin/sh$#/new/path#' /etc/passwd <-- Single quotes around the -e block avoid having to escape for the shell. If you're uncomfortable editing the passwd file in place, you can use perl -pi.old -e 's#/bin/sh$#/new/path#' /etc/passwd, which will put the original in /etc/passwd.old.

I posted the long way in case you wanted to do more with the data than sed can do ...

(Note that the '?'s from the last poster and the '#'s in this are completely interchangeable, but '?' is a regex reserved character and makes the search string hideous to read).

If it's jjust one user, why not just vipw and be done with it?

phlx · 12-03-2004, 01:59 PM

well here is what I whipped up, but its not working..

#!/usr/local/bin/perl -w
use strict;
open FILEHANDLE, 'passwd' or die $!;

while (<FILEHANDLE>) {
my $line =~ s{/bin/bash\n}{/bin/csh\n};
push(@output, $line);
}

close FILEHANDLE;

also, I'm not sure, but is there another way I could do this directily to STDout instead of loading it all into @output to print? (that is what print does correct ?)

as of right now my one liner perl proggy is

#!/usr/local/bin/perl -w

system perl => -pi => -e => 's#/bin/bash$#/bin/csh#g' => passwd;

sigsegv · 12-03-2004, 03:01 PM

Code:

#!/usr/local/bin/perl -w

use strict;
open FILEHANDLE, 'passwd' or die $!;

while (<FILEHANDLE>) {
        my $line =~ s#/bin/bash\n#/bin/csh\n#;
        print $line;
}

close FILEHANDLE;

That should fix that, and print straight to STDOUT.

If you're looking for a one liner, it would look like this:

Code:

(to STDOUT)
perl -pe 's#/bin/bash\n#/bin/sh\n#' /etc/passwd

(to /etc/passwd)
perl -i -pe 's#/bin/bash\n#/bin/sh\n#' /etc/passwd