unziping a file in python..

[Gabriel Genellina]

En Mon, 02 Mar 2009 13:09:05 -0200, Luis Zarrabeitia <kyrie at uh.cu>  
escribió:
> Quoting MRAB <google at mrabarnett.plus.com>:
>> Steven D'Aprano wrote:
>> >
>> You might want to specify an output folder (and the data might be binary
>> too):
>>
>> zf = zipfile.ZipFile('Archive.zip')
>> for name in zf.namelist():
>>      open(os.path.join(output_folder, name), 'wb').write(zf.read(name))
>>
>
> Question here... wouldn't you also need to create all the intermediate  
> folders
> from "output_folder" to "output_folder/name" (assuming that 'name' has  
> several
> path parts in it)? Is there any elegant way to do it?

Perhaps it would be easier to grab a copy of the 2.6 sources. extract()  
and extractall() are not so complex, but "the devil is in the details"

> (is there any way to make os.mkdir behave like mkdir -p?)

if not os.path.isdir(d):
   os.makedirs(d)

-- 
Gabriel Genellina