En Tue, 03 Feb 2009 05:31:24 -0200, Brandon Taylor
<btaylordes...@gmail.com> escribió:
I'm having an issue specifying the path for extracting files from
a .zip archive. In my method, I have:
zip_file.extract(zip_name + '/' + thumbnail_image, thumbnail_path)
What is happening is that the extract method is creating a folder with
the name of 'zip_name' and extracting the files to it. Example:
extract will create all directories in member name. Use open instead:
with zip_file.open(zip_name + '/' + thumbnail_image) as source:
with open(os.path.join(thumbnail_path, thumbnail_image), "wb") as target:
shutil.copyfileobj(source, target)
(untested)
--
Gabriel Genellina
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