You'll have to create the directories separately from the rename() because rename() doesn't do that. Read perldoc File::Path. Look for the mkpath routine.

Note that in addition, rename doesn't (typically) span filesystems; so you end up needing File::Copy::move or equivalent, too.

Indeed!

duff

Thanks guys, I had suspected that rename didn't support that but didn't know what alternative to seek. Now I'm stuck tryiing to make a regex that will chop the filename off of a directory string like c:\bla\did\dy\blah.txt to be c:\bla\did\dy. Can you answer this one as fast?

Currently I am splitting the string on "\" and reconstructing it without the last array element. Surely there is a better way.

Currently I am splitting the string on "\" and reconstructing it without the last array element. Surely there is a better way.

Indeed there is, use File::Basename, and thy foot shall step upon the path of goodly ways.

### ... sample snippet ... 
my ($basename,$path,$extension) = 
     File::Basename::fileparse($fullpath,qw(.txt));
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Update: Hmmm, actually, if it doesn't do the "create directories" bit -- the destination doesn't exist at all -- then you may need to also muck about with File::Path::mkpath.

You can make use of the File::Basename as scooterm suggested.

$path = c:\bla\did\dy\blah.txt ;
$filename = basename($path);
$dirname = dirname($path);
print "filename:$filename\ndirecory:$dirname";
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output:

filename:blah.txt
directory:c:\bla\did\dy
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... but note that this method does not work if your path is specified using forward-slash ("/") as the pathstep delimiter.