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

How can I adapt this bit of code; 
my $size = 0;
$size += (-f $_ ? -s _ : 0) for (<*>);
print $size;

[download]
so that instead of <*> i can pass on a remote share or directory name?

like this: \\Server\Share_Name or \\Server\DriveLetter\FolderName.

Since I have not tried this code yet so I don't know how fast it is, but I have tried these 2 methods 

1) ($Total)=( (qx{dir /s "$UncSharePath"})[-2]=~ /([\d,]+) bytes/ )

2) find(sub { $total += -s }, $dir);

[download]
And they are very slow -unusable really in real life, ok on small servers thu.
Does any one know how can I extract the sizes of a remote share or folder directly off the OS? Surely automating this process in Perl should be faster than doing it manually, i.e. by right clicking on a folder and selecting the properties, which seems to be the fastest method, but unfortunaly I have a list of servers to go through...
Thanks for your help Guys.

Replies are listed 'Best First'.
Re: Faster way to obtain a remote share or folder size
by BrowserUk (Patriarch) on Jul 01, 2003 at 16:42 UTC

    Take a look at this thread Find A Share size in Win2k. There is a method there by jsprat that is blindingly fast if the shared directory is the root of the remote drive.

    Examine what is said, not who speaks.
    "Efficiency is intelligent laziness." -David Dunham
    "When I'm working on a problem, I never think about beauty. I think only how to solve the problem. But when I have finished, if the solution is not beautiful, I know it is wrong." -Richard Buckminster Fuller


[reply]
Re: Faster way to obtain a remote share or folder size
by TomDLux (Vicar) on Jul 01, 2003 at 16:55 UTC

    Your headline asks for a faster way, but the message asks how you can get the size of a directory other than the one where you are.

    Of course, I'm writing on Linux, but the routine should work on all platforms.

    my $dirname = "/usr/local/bin";
my $size = get_dir_size( $dirname );

# Perl Cookbook recipe 9.5
#
sub get_dir_size {
    return unless @_ && -e $_[0] && -d _ && -r _ ;
    my ( $dir ) = @_;
    my ( $file, $size ); 
    opendir DIR, $dir or die( "Could not open directory '$dir'\n" );
    while ( defined ( $file = readdir( DIR ))) {
        $size += -s "$dir/$file";
    }
    closedir DIR;
    return $size;
}

    [download]

    Of course, the easy answer, especially if you're interested in subdirectories, too, is:

    $size = `du -s -k $dir`;

    [download]

    --
    TTTATCGGTCGTTATATAGATGTTTGCA

[reply]
[d/l]
[select]
      Thanks guys, I 'll try out your suggestions...

      And I'll try and be more precise with wording of my posts :)

      Cheers.
[reply]

Back to Seekers of Perl Wisdom