Hi:
I am starting a new thread based of an older thread just because there was a
lot of different things that were requested for and it had gotten a bit
confusing.
I got a lot of help from all you experts to write the below code that takes
an argument path C:\build\Sample\NewDir that contains--
C:\build\Sample\NewDir\File1.txt
C:\build\Sample\NewDir\File2.txt
C:\build\Sample\NewDir\NewSubDirectory
C:\build\Sample\NewDir\NewSubDirectory\11.txt
and prints out the following into the output file --
File1.txt
File2.txt
NewSubDirectory
use strict;
use warnings;
my $path = $ARGV[0];
opendir DIR, $path or die "Can't open $path: $!";
my @new = grep { $_ ne "." and $_ ne ".." } readdir DIR;
closedir DIR;
open FILE,">>c:/buildlist2.txt";
print FILE "$_\n" foreach @new;
close FILE;
Can I modify the above code so that no directory name is printed in the
ouputfile but only filenames are printed. ie File1.txt and File2.txt are
printed in the output file without the NewSubDirectory printed in it? I am
looking for some way that before getting the entries into the new array, i
can remove the entries that stand for directory names.
I tried using find(sub {push @new, $_ if -f}, $path); but that prints all
the filenames under NewDir as well as NewSubDirectory together which is not
what i need. For getting files under NewSubDirectory, I will issue a
separate command with argument path as
C:\build\Sample\NewDir\NewSubDirectory and hence this command does not work
since it clubs files and directories and files within sub-directories.
I tried using find(sub {push @new, $File::Find::name}, $path); but that
prints the directories and files along with their complete/absolute paths,
which is not what i need.
Thanks in advance! -Nishi.
