Extracting file from zip archive in Python 2.6.1

[Brandon Taylor]

On Feb 3, 9:45 am, "Gabriel Genellina" <gagsl-... at yahoo.com.ar> wrote:
> En Tue, 03 Feb 2009 05:31:24 -0200, Brandon Taylor  
> <btaylordes... at 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