Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Dear monks,

I have a directory containing many new directories all of which contain a file called 'mlc'. I want to be able to read the contents of each 'mlc' file from all directories (115) from with a perl script.

Is this possible? I also want to be able to relate the contents of each file to the name of the directory that it came from.

As there are so many directories I really don't want to hard code it.

All I have so far is listing the files in the directory 'new'.

'new' then contains the 115 directories all containing a file 'mlc' (which all have different contents). All of the directories within new are named the same apart from a number at the end, e.g. dir0 -> dir114 

opendir (NEW, "new/") or die "unable to open directory new $!";

my @dir_files = readdir(NEW);
print "@dir_files\n";

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Replies are listed 'Best First'.
Re: opening directories in perl
by adrianh (Chancellor) on Jul 03, 2005 at 20:33 UTC

    Take a look at File::Find::Rule, which is much easier than File::Find in most instances. You'd want something like (untested):

    use File::Find::Rule;
my @mlc_files = File::Find::Rule->file->name( 'mlc' )->in( 'new/' );

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Re: opening directories in perl
by TedPride (Priest) on Jul 03, 2005 at 19:35 UTC
    I tested the following on my Mac, so it uses : instead of /, but it should provide the basis for what you need: 
    use strict;
use warnings;

my ($dir, $handle, %mlc, @files, @spider);

my $fname = "mlc";
# Main directory to start spidering...
push @spider, ':';

while ($dir = shift @spider) {
    if (-e "$dir$fname") {
        open($handle, "$dir$fname");
        $mlc{$dir} = join '', <$handle>;
        close($handle);
    }
    opendir($handle, $dir);
    @files = readdir($handle);
    closedir($handle);
    for (@files) {
        push @spider, "$dir$_:" if -d "$dir$_";
    }
}

for (sort keys %mlc) {
    print "$_\n$mlc{$_}\n\n";
}

    [download]
    Of course, you could make things much simpler and just cycle through all the directories with numbers 0 through x, where x is the maximum possible value: 
    use strict;
use warnings;

my (%mlc, $handle);
for (0..200) {
    if (-e ":dir$_:mlc") {
        open($handle, ":dir$_:mlc");
        $mlc{"dir$_"} = join '', <$handle>;
        close($handle);
    }
}

for (sort keys %mlc) {
    print "$_\n$mlc{$_}\n\n";
}

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Re: opening directories in perl
by mifflin (Curate) on Jul 03, 2005 at 18:33 UTC
    Take a look at the File::Find module.
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Re: opening directories in perl
by trammell (Priest) on Jul 03, 2005 at 18:33 UTC
    You can certainly brew up your own solution with opendir(); one other (possibly easier) solution would be to use module File::Find.
[reply]
      I'm having some real problems with this File::Find module!

      I have read the relevent docs but am really unsure what to but in the wanted subroutine in order to find the files I want. The files I want to get are new/something_here/mlc 

      use File::Find;

my $dir = ('new/');
find (\&wanted, $dir);

# this bascially lists all the directories, how can I get it to look f
+or my 'mlc' files?

sub wanted      {
                print $File::Find::dir;

}

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        In the documentation I find this snippet: 
        The wanted() function does whatever verifications you
want.  $File::Find::dir contains the current directory
name, and $_ the current filename within that directory.
$File::Find::name contains the complete pathname to the
file.
        So I'd try this to start (untested): 
        my @mlc;
sub wanted {
    return unless $_ eq 'mlc';
    push @mlc, $File::Find::name;
}
find(\&wanted, '.');
foreach my $file (@mlc) {
    open(my $fh, $file) or die "Can't open '$file': $!";
    # per-file code goes here
}

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