Hi Davis,
Thanks for your quick resonse
If you look at the dir variable, you'll see a / in front of
it. And the load_file always requires a fully qualified
pathname.
---------------------------
Dr. Mark Ceulemans
Senior Consultant
IT Masters, Belgium
| [reply] |
Hi there,
The call to readdir will not return the full path - here's an example I've just tried:
#!/usr/bin/perl -w
use strict;
my $dir = "/scratch";
opendir(DIR, $dir)
or die "Unable to opendir $dir: $!\n";
foreach my $file (readdir DIR) {
print "file is at: $file\n";
}
closedir(DIR);
That code will only print the file's paths relative to /scratch - in order to get the fully qualified path, I'd need to prepend the value of $dir.
Note that this may still be barking up the wrong tree ;-)
davis
Is this going out live?
No, Homer, very few cartoons are broadcast live - it's a terrible strain on the animator's wrist
| [reply]
[d/l] |
Hi Davis,
Thanks for your quick response
If you look at the dir variable, you'll see a / in front of
it. And the load_file always requires a fully qualified
pathname.
---------------------------
Dr. Mark Ceulemans
Senior Consultant
IT Masters, Belgium
| [reply] |