> On Tuesday, Nov 18, 2003, at 09:20 US/Pacific, Dan Muey wrote:
>
> > Howdy group.
> >
> > I have a need to get the size of a directory.
> > I could just execute unix's du command (my $sz = `du -sh
> /usr`;) but
> > was wondering if there's a better way to do it that is more
> portable.
> > I looked on cpan but didn't see anythign that jumped out at me.
> >
>
> a part of the problem is that the 'du' command
> is basically implemented differently between the
> BSD and SYSV styles of unix. One strategy that
> I have adopted when I am looking at a 'common'(HA!)
> unix utility that I would like to use, and
> in your illustration, the 'flags' to it, that
> are problematic, I create a wrapper class.
>
> a quicky way to think about your problem would
> be my WhackJob for dealing with my 'favorite'
> set of flags to the 'ps' command:
>
> cf
> <http://www.wetware.com/drieux/CS/Proj/PID/>
>
> To be honest, I use something like that, but more
> cleaned up in a project I have. The trick of
> course is getting 'illustrations' from all of the
> OS's that you will want to be 'portable' with.
>
Thanks. What I really want to do is avoid system commands completely.
I could always traverse the directory adding the size of each file
together and do that all in Perl with no system commands but I figured
there'd already be a Module or something like this, so I want to do it
that way. Somthign like:
my $sz = $module->size("/home/joemama");
# return sixze of file or of entire directory if specified file is a directory
> HTH.
>
> ciao
> drieux
--
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
