Get filesize with -s in Perl

Question

I am trying to find the size of a file using the -s operator. It looks like this:

my $filesz = -s $filename

I tried lots of various way, but it can not get this size.
However, if I give static content instead of filename, it works fine

For example:

$filesz = -s "/tmp/abc.txt"

This works fine.

I tried adding " in the filename, it didn't work. I removed \n from filename using chomp, but the problem remains the same. What's wrong here?

Answer 1

-s $filename works just fine; the only conclusion is that there's no file with the name contained in $filename. Take a very close look at the contents of $filename, and make sure that your working directory is what you think it is.

Answer 2

As hobbs says, the most likely explanation is that $filename doesn't contain what you think it does.

Based on previous experience, I'd go further than that and hesitate a guess that $filename has a newline character at the end of it. Are you reading the value in $filename from a file or from user input?

Answer 3

i think you should try this your getting from a readdir:

#!/usr/bin/perl -w

use strict;
use warnings;

my $filedir = "/documents/sample/file.txt";


opendir(my $dir, $filedir) or die "Could not load directory";
my @read_dir = sort grep {!-d}readdir($dir);
foreach my $fileInDir(@read_dir)
{

    my $currentDestination = "$filedir/$fileInDir";
    my $filesize = -s $currentDestination;

    if(-z $currentDestination)
    {
       print "Zero -  Filename: $fileInDir Size: $filesize\n";
    }
    else
    {
        print "Non - Zero Filename: $fileInDir Size: $filesize\n";
    }

}