How to extract the column data from multidimensional array in perl - perl

Question

ls -l
-rw-r--r-- 1 angus angus 0 2013-08-16 01:33 copy.pl
-rw-r--r-- 1 angus angus 1931 2013-08-16 08:27 copy.txt
-rw-r--r-- 1 angus angus 492 2013-08-16 03:15 ex.txt
-rw-r--r-- 1 angus angus 25 2013-08-16 09:07 hello.txt
-rw-r--r-- 1 angus angus 98 2013-08-16 09:05 hi.txt
I need only the read, write , access data as well as the file name.
#! /usr/bin/perl -w
#list = `ls -l`;
$index = 0;
#print "#list\n";
for(#list){
($access) = split(/[\s+]/,$_);
print "$access\n";
($data) = split(/pl+/,$_);
print "$data";
#array1 = ($data,$access);
}
print "#array1\n"
I have written this code to extract the read,write,access permission details and the file name corresponding to it.
I couldn't extract the filename which is the last column.

Answer 1

Check perl stat http://perldoc.perl.org/functions/stat.html
It's more robust and efficient than calling external ls command,
use File::stat;
$sb = stat($filename);
printf "File is %s, size is %s, perm %04o, mtime %s\n",
$filename, $sb->size, $sb->mode & 07777,
scalar localtime $sb->mtime;

Answer 2

I think you have an error in line number 8 of your script. You are trying to split the line using the string "pl" as a delimiter which will only match the first line of your input and will not give you what I think you want.
I believe you should just split the whole line on white space and assign just the columns you want (number 1 and 8 in this case).
change your loop for this:
for my $filename (#list){
chomp($filename);
my ($access, $data) = (split(/\s+/, $filename))[0, 7]; #use a slice to get only the columns you want.
print "$access $data\n";
}
Note: mpapec suggestion to use Stat would be better. I just wanted to let you know why your code is not working.