[SOLVED] passing perl script argument to an outside program

bspears1 · 11-14-2012, 03:48 PM

I have a program called checkuser, called from the shell with this syntax:
checkuser <user>@<domain> dsl

I am attempting to write a perl script that takes a username as an argument, and gets info from checkuser. The output is <100 lines, so I just want to pull it as a list of lines, straight to memory.

I hope the following will further suggest what I am attempting (though it still seems incorrect..):
my @checkuserpull = `checkuser $1@domain dsl`

As always, any assistance is greatly appreciated

Birei · 11-14-2012, 04:09 PM

Hi bspears1,

Arguments are saved in the array @ARGV. So use $ARGV[0] for first parameter, $ARGV[1] for the second one and so on.

Code:

my @checkuserpull = `checkuser ${ARGV[0]}@domain dsl`

bspears1 · 11-14-2012, 04:30 PM

@Birei,

So I tried:

Code:

#!/usr/bin/perl
my @checkuserpull = `checkuser ${ARGV[0]}@domain dsl`;

I then did:
./testscript test.account6

It replied:
checkuser: invalid user name

However, if I do the following, from the shell, it works just fine:
checkuser test.account6@domain dsl

So.. something still isn't being passed correctly.

Birei · 11-14-2012, 04:49 PM

Perl assumes that @domain is an array. You must escape it:

Code:

my @checkuserpull = `checkuser ${ARGV[0]}\@domain dsl`

bspears1 · 11-14-2012, 04:56 PM

Sweet

Thank you, Birei!

Sergei Steshenko · 11-14-2012, 07:09 PM

Quote:

Originally Posted by bspears1

Sweet

Thank you, Birei!

You are doing it methodologically wrong - regardless of language. Your fundamental mistake is the assumption you'll make no mistake.

If you change that assumption,. you'll start writing your code in the manner which allows you to easily check inputs and outputs of intermediate steps.

But it appears you haven't used

Code:

use strict;
use warnings;

in the beginning of your script in the first place - which would have pointed you to the mistake immediately.

Now, back to the wrong assumption - you have to write your code using the following idiom:

Code:

my $cmd = "checkuser ${ARGV[0]}@domain dsl"; # this line is actually wrong
warn "about to get stdout from $cmd command line"; # comment this line out when your code works
my @checkuserpull = `$cmd`;

.

bspears1 · 11-14-2012, 07:29 PM

@Sergei

You assume that I made an assumption about my infallibility. In fact, I normally expect to mess it up, and find mistakes. I did not use

Code:

use strict;
use warnings;

because I am very new, and unaware of such options. They will be added, for sure, since I still have quite a bit to do.

Birei's alterations to my example made it possible to successfully get the information I need. If I have chosen a poor method, perhaps you could suggest a better way? It would be appreciated

Sergei Steshenko · 11-15-2012, 12:27 PM

Quote:

Originally Posted by bspears1

@Sergei

You assume that I made an assumption about my infallibility. In fact, I normally expect to mess it up, and find mistakes. I did not use

Code:

use strict;
use warnings;

because I am very new, and unaware of such options. They will be added, for sure, since I still have quite a bit to do.

Birei's alterations to my example made it possible to successfully get the information I need. If I have chosen a poor method, perhaps you could suggest a better way? It would be appreciated

The method of getting the needed is OK, non-checking of actual command line to be executed is not.

My assumption of me screwing up command lines is carved in stone, code code that is:

Code:

cat -n system_wrapper.prl
     1  ###############################
     2  # written by Sergei Steshenko #
     3  ###############################
     4
     5  use strict;
     6
     7  sub
     8    {
     9    my ($info_marker, $command)=@_;
    10
    11    warn "$info_marker executing ==> ",$command,"\n";
    12    # ^
    13    # | - it should be to STDERR, not STDOUT, UNIX(R) traditions should be respected.
    14    system($command);
    15    };

cat -n backticks_wrapper.prl
     1  ###############################
     2  # written by Sergei Steshenko #
     3  ###############################
     4
     5  use strict;
     6
     7  sub
     8    {
     9    my ($info_marker, $warning_marker, $command)=@_;
    10
    11    warn "$info_marker going to grab STDOUT from '$command' command\n";
    12    my $stdout=`$command`;
    13    if($?)
    14      {
    15      warn "$warning_marker '$command' command returned non-zero status: \$?=$?";
    16      }
    17
    18    $stdout; # the returned value
    19    };

.