could you please tell me why I am getting the error below?

The 1st open in your while loop tries to open the directory named '.'.

I believe you only want to open files. To do so, use the -f filetest like this:

use strict;
use warnings;

my $dir = "C:/move/output/test/";
opendir DIR, $dir or die "Unable to open $dir: $!";
while (my $filename=readdir DIR)
{
    my $file = "$dir/$filename";
    next unless -f $file;
    open my $fh, '<', $file or die "Unable to open $file: $!";
[download]

Thank you toolic for your great help although I need to read more about "unless". I am not sure I understand it totally. I tested the script and for some reason it only reads the first text files in the directory (they are all text files). How I can generate a text files in the output for every text files that it reads in?

#!/usr/bin/perl
use strict;
use warnings;

my $dir="C:/move/output/test";
opendir DIR, $dir or die "Unable to open $dir: $!";

while (my $filename=readdir DIR)
{
my $file="$dir/$filename";
next unless -f $file;   
open my $fh, '<', $file or die "Unable to open $filename: $!";

open OUT,">C:/move/output/test/out.txt";
select OUT;
my (variables);
{ my loop}
close $fh;
}
[download]

The code you posted will open every file that exists and passes the -f (entry is plain file) file test.

Unfortunately, it will also open the same output file on each pass of the while loop -- with write permission (not append). Ie, every pass through the loop creates a completely new c:/move/output/test/out.txt file. I suspect this is the reason you believe you are only finding a single file.

To append to the file on each pass, change your open command from open OUT,">C:/move/output/test/out.txt"; to open OUT,">>C:/move/output/test/out.txt"; You should also consider checking the return value of open.

Yes it is obvious if you read the documentation for readdir.

Thank you J.C. If you mean the "/" after test, then it was a typo. I am not sure what you mean by ignoring it