mh53j_fe has asked for the wisdom of the Perl Monks concerning the following question:

I want to write a small Perl script that will output to a browser the contents of a directory in the format of the command line function ls -lt. I only want to see those files in the directory that end with the following: *.log, *.txt, *.sql, *.err, *.par, *.csv. I want to use this file to troubleshoot report problems on our production web server. (Developers do not have access to production at my company.)

Here is what I have written so far: 

#!/usr/local/bin/perl5.00506
use CGI;
$query = new CGI;
$query_string = $query->param('DIR');
print "Content-type: text/plain", "\n\
print `ls -lt $query_string`;

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I want do something like: 
print `ls -lt *.sql *.txt *.log .. $query_string`;

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However, this does not work. Any help that you can provide will be highly appreciated. Regards,

Dave

Replies are listed 'Best First'.
Re: Outputting contents of ls -lt
by marto (Cardinal) on May 04, 2005 at 16:29 UTC
    Hi Dave, 

    #!/usr/bin/perl
use strict;
use CGI;
my $query = new CGI; 
my $filter = "*.pl *.txt";
print "Content-type: text/plain"; 
my $lsoutput = `ls -lt $filter`;
print "\n$lsoutput";

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    I have included use strict; and a variable called $filter which, in the example provided, returns the output for the following command ls -lt *.pl *.txt

    Hope this helps

    marto
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Re: Outputting contents of ls -lt
by marto (Cardinal) on May 04, 2005 at 17:39 UTC
    Hi again !

    About the use strict; thing.
    This is a good node (thanks runrig) to have a read at to introduce why to include use strict;, just incase you did not know already.

    Cheers,

    marto
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Re: Outputting contents of ls -lt
by tlm (Prior) on May 04, 2005 at 19:08 UTC

    Why exactly did print `ls -lt ...` fail? Also, what information do you need from the output of ls -lt? Some it can be obtained quite easily by other means (e.g. -s, -M) than don't require invoking ls. Here's a quick, poor-man's ls: 

    use strict;
use warnings;
my $t   = time;
my $spd = 24*60*60;

printf "%d\t%s\t%s\n", -s, scalar(localtime($t+$spd*-M)), $_
  for <*.{log,txt,sql}>;

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    BTW, it is probably not a good idea to stick an input string right into backticks like you have in your sample code; what if someone gave you an input string like '; rm -rf * *.*' ?

    the lowliest monk

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Re: Outputting contents of ls -lt
by TedPride (Priest) on May 04, 2005 at 16:26 UTC
    Put the return value from ls into a variable, and filter the file names for the extensions you want to allow. It's better to do it this way than filtering through ls, imho.

    EDIT: Here's some code: 

    #!/usr/local/bin/perl

use strict;
use warnings;

my $base = '/u/web/user/';
my @allow = qw/log txt sql err par csv/;
my $e;

print "Content-type: text/plain\n\n";
for (split /\n/, `ls -lt $base`) {
    for $e (@allow) {
        if (m/\.$e$/) {
            print "$_\n"; last;
        }
    }
}

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    The advantage of doing it this way is that your code is easy to modify to work with rejected lines, should you decide you want to in the future.
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