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Print the name of files in the directory using Perl

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# 1  
Old 03-13-2012
Print the name of files in the directory using Perl

Hi All,
I am stuck with a problem and i want your help, what i want to do is

I have a directory name dircome
and there are 6 files present in this directory, the name of the files are d1,d2,d3,d4,d5,d6.

The Perl script print the files name, means the output should be
# 2  
Old 03-13-2012
Try like...
ls -l *.*| awk -F ' ' '{print $9}'

# 3  
Old 03-13-2012
Use the function system.
#! perl
system("ls <path to directory>");


Last edited by pludi; 03-13-2012 at 07:50 AM..
# 4  
Old 03-13-2012
Above command not working.. Please help me.
# 5  
Old 03-13-2012

Try this one,
#! /usr/bin/perl
use strict;
my $mDir="/home/cf/ranga1";
opendir(DIR1,"$mDir") or die "unable to open dir $mDir";
my @mRes=grep{/^[^.]/} readdir(DIR1);
print "@mRes\n";


Last edited by rangarasan; 03-14-2012 at 05:18 AM..
# 6  
Old 03-13-2012
A one-liner using only Perl (unlike the others presented so far):
perl -e 'print join "\n",grep { -f } <*>'

  • <*> is a glob that expands to all entries in the current directory
  • grep is a function that works pretty much like the shell utility. It expects a statement block returning true or false values on the parameters.
  • The -f is a shorthand for testing if a parameter (here it's the anonymous automatic variable $_) is a regular file or not
  • And finally it all gets nicely formatted by join and printed
# 7  
Old 03-13-2012
If you are running a Perl script there is no need to spawn a shell to do this, Perl has a number of ways (as always) of doing this, you can use a glob to get the values
print join("\n",glob("$dir_name/*")),"\n";

Or you could open the directory and read the contents.
opendir(my $directory,"$dir_name"); 
print join("\n", grep {! /^\./} sort(readdir $directory));

The glob is obviously easier to use, however if you need to do more complex processing than just printing the filenames the opendir, readdir , closedir approach gives you more flexibility.

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