QUOTE(Synthetic Soul @ Sep 30 2005, 07:44 AM)
Actually, mine does too.
I'm sorry, I didn't test the string kockroach posted. I assumed it must work as he uses it. Never assume...
I've now tested and it
does work in 0.8.3, just
not in 0.9.
I don't think you can use $directory() in 0.9 in this situation. I tested the string as my
copy command, and it works as expected.
"WAVs\$directory(%path%)\%filename%" does not work, but "WAVs\%filename%\%filename%" and "WAVs\%path%\%filename%" do.
The results are not quite what I was expecting! Ideally, the subdirectory structure would be replicated as a whole. Instead, each output file is placed in its own, uniquely named directory.
When I use "WAVs\%path%\%filename%" I get the following
subdirectories created in "WAVs" (yes, these are
directories even though they have the .flac extension; each contains the respective output .wav file):
CODE
I__Beatles - mono collection (EMI-Parlophone BM1)_UK 45 Collection_1962-10-05 single (Parlophone 45-R 4949)_01 Love Me Do.flac
I__Beatles - mono collection (EMI-Parlophone BM1)_UK 45 Collection_1962-10-05 single (Parlophone 45-R 4949)_02 PS I Love You.flac
I__Beatles - mono collection (EMI-Parlophone BM1)_UK 45 Collection_1963-01-11 single (Parlophone 45-R 4983)_01 Please Please Me.flac
I__Beatles - mono collection (EMI-Parlophone BM1)_UK 45 Collection_1963-01-11 single (Parlophone 45-R 4983)_02 Ask Me Why.flac
... and switching to "WAVs\%filename%\%filename%" yields a similar result, save that the subdirectory names are simply the same as the file names or the original .flac files.
Any thoughts?
- M.
Edit: Cleaned up post a bit.