CODEWITHSUNDEEP
perlpipe
Pinky Sharma
Pinky Sharma

Reputation: 31

how to use a variable in a command used for open along with (-|) pipe output where the filename is interpreted as a command that pipes output to us

How can I use a variable in a command used for open along with (-|) pipe output where the filename is interpreted as a command that pipes output to us.

$cmd = 'ps -elf';
open( my $fh, "-|",$cmd  ) || die( "$cmd failed: $! " );

Here I want $cmd = 'ps $myOptions'; where $myOptions will be set to the required options lets say for example $myOptions = "-elf"

How can this be done?

Upvotes: 2

Views: 544

Answers (3)

mirod
mirod

Reputation: 16161

Double quoting the pipe works for me:

#!/usr/bin/perl

use strict;
use warnings;

my $cmd = 'ps';
my $opt = "-elf";
open( my $fh, "-|", "$cmd $opt"  ) || die( "$cmd failed: $! " );

while( <$fh>) { print "line $.: $_"; }

Also working: "ps $opt", join( ' ', $cmd, $opt), $cmd . ' ' . $optand probably many other ways. You just have to make sure that the 3rd argument to open is a string, with the proper contentps -elf`. For this you have to make sure you interpolate the variables (ie no single quote), and you don't end up with a list instead of a string (ie concatenate or use variables between double quotes).

Upvotes: 1

SumitKhanna
SumitKhanna

Reputation: 44

You can use the string concatenation for this as $cmd = "ps ".$myOptions;

Upvotes: 2

user149341
user149341

Reputation:

If you want to specify the command as a single string (e.g, if your options may actually consist of several arguments):

my $cmd = "ps $myOptions";
open my $fh, "$cmd |" or die "$cmd failed: $!";

If you want to specify it as two strings (e.g, if $myOptions should always be treated as a single argument):

open my $fh, "-|", "ps", $myOptions or die "ps $myOptions failed: $!";

Upvotes: 0

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