On Feb 3, 1:16 pm, Brandon Taylor <btaylordes...@gmail.com> wrote: > On Feb 3, 9:45 am, "Gabriel Genellina" <gagsl-...@yahoo.com.ar> wrote: > > > > > 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 > > Hi Gabriel, > > Thank you for the code sample. I figured I was going to have to use > 'open', but I completely forgot about the 'with' statement. I was > trying to figure out how to get access to the file object in the zip > without having to loop over all of the items, and 'with' will allow me > to do just that. > > I'll give it a shot when I get home this evening and post my results. > > Kind regards, > Brandon
Ok, the first thing I needed to do was add: from __future__ import with_statement at the beginning of my file but: 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) Returns an error on the first line: ZipExtFile instance has no attribute '__exit__' Googling this error message is turning up nothing, and there's no mention of the exception in the docs. Any thoughts? TIA, Brandon -- http://mail.python.org/mailman/listinfo/python-list