One problem with deluge.metafile.dummy()
Posted: Mon Jun 22, 2009 9:17 am
I was try to create a torrent file from metafile libary.
But there have one problem with argument.
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Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/var/lib/python-support/python2.6/deluge/metafile.py", line 96, in make_meta_file
info = makeinfo(path, piece_length, progress, name, content_type, private)
File "/var/lib/python-support/python2.6/deluge/metafile.py", line 216, in makeinfo
progress(piece_count, num_pieces)
TypeError: dummy() takes exactly 1 argument (2 given
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I'm check metafile for dummy,
But there have one problem with argument.
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Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/var/lib/python-support/python2.6/deluge/metafile.py", line 96, in make_meta_file
info = makeinfo(path, piece_length, progress, name, content_type, private)
File "/var/lib/python-support/python2.6/deluge/metafile.py", line 216, in makeinfo
progress(piece_count, num_pieces)
TypeError: dummy() takes exactly 1 argument (2 given
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I'm check metafile for dummy,
It should be two input value.def dummy(v):
pass
make_meta_file(....,progress=dummy,)
makeinfo(...
progress(piece_count, num_pieces)
def dummy(v,n)
pass