Shlomi/Uri,
what do you mean by 'It's a good idea not to use "select" on filehandles like
that because it willaffect the default filehandle permanently.'?
I am using perl 5.14
ty.
________________________________
From: Shlomi Fish <shlo...@shlomifish.org>
To: perl list <beginners@perl.org>
Sent: Friday, June 28, 2013 4:48 AM
Subject: Re: printing content of a pipe
Hi Rajeev,
see below for some comments on your code.
On Thu, 27 Jun 2013 15:07:50 -0700 (PDT)
Rajeev Prasad <rp.ne...@yahoo.com> wrote:
> in the below code I am not able to print anything except whatever is in the
> $pty. I want to print LINE_START: in the beginning of each line of the
> output, but it does not print that, it only prints what the output of the
> exceuted command produced. why is that so?
>
> while(<$pty>) {
>
> print "LINE_START: $. $_";
> }
>
>
>
> I am suspecting it has something to do with select STDOUT; $| = 1;, but i
> dont know what.
>
> complete code:
>
> #!/usr/bin/perl
> use strict;
> use warnings;
The strict and warnings are a good idea.
> use Net::OpenSSH;
> use Expect;
>
> select STDOUT; $| = 1;
> select STDERR; $| = 1;
It's a good idea not to use "select" on filehandles like that because it will
affect the default filehandle permanently. Instead do:
<CODE>
use IO::Handle;
STDOUT->autoflush(1);
STDERR->autoflush(1);
</CODE>
>
> my $password = $ARGV[0];
> my $target = $ARGV[1];
> my $runas = $ARGV[2];
> my $cmd = $ARGV[3];
This uses positional indexes and is subject to error. See:
http://perl-begin.org/tutorials/bad-elements/#subroutine-arguments
(my own link).
> my $timeout = 20;
> my $debug = 0;
>
> my $ssh = Net::OpenSSH->new($target,
> user => 'abcd',
> password => $password,
> master_opts => [-o => 'StrictHostKeyChecking=no',
> -o => 'ConnectTimeout 60']
> );
The formatting here is inconsistent.
> my $sshExitCode = $ssh->error;
> if ($sshExitCode eq "0") {
> my $ostype= $ssh->capture("uname -s");chomp($ostype);
> my $sudo_path=$ssh->capture("which sudo");chomp($sudo_path);
> my $su_path=$ssh->capture("which su"); chomp($su_path);
1. You've placed two statements on the same line three times.
2. You may wish to extract a subroutine or a closure for doing that:
<CODE>
my $read_str = sub {
my ($cmd) = @_;
my $ret = $ssh->capture($cmd);
chomp($ret);
return $ret;
};
</CODE>
Also see what Uri said.
Regards,
Shlomi Fish
--
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