Could zipfile module process the zip data in memory?
kelvin.you at gmail.com
Sun Apr 29 14:16:37 CEST 2007
On Apr 29, 7:37 pm, "Daniel Nogradi" <nogr... at gmail.com> wrote:
> > I made a C/S network program, the client receive the zip file from the
> > server, and read the data into a variable. how could I process the
> > zipfile directly without saving it into file.
> > In the document of the zipfile module, I note that it mentions the
> > file-like object? what does it mean?
> > class ZipFile( file[, mode[, compression[, allowZip64]]])
> > Open a ZIP file, where file can be either a path to a file (a
> > string) or a file-like object.
> Yes it is possible to process the content of the zipfile without
> saving every file:
> from zipfile import ZipFile
> from StringIO import StringIO
> zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' )
> for name in zipp.namelist( ):
> content = zipp.read( name )
> s = StringIO( )
> s.write( content )
> # now the file 'name' is in 's' (in memory)
> # you can process it further
> # ............
> s.close( )
> zipp.close( )
OK, I see, thank you!
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