Could zipfile module process the zip data in memory?

[人言落日是天涯，望极天涯不见家]

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:
>
> [untested]
>
>         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( )
>
> HTH,
> Daniel
Thanks!
Maybe my poor english makes you confusion:-). The client receive the
zip file data from the server, and keep these data as a variable, not
as a file in harddisk. such as "zipFileData", but the first argument
of the "ZipFile" is filename. I would like to let the ZipFile() open
the file from "zipFileData" directly but not file in harddisk

zipp = ZipFile( this_is_the_zip_file_from_your_server, 'r' )
                              ^ I don't have this file, all its data
is in a variable.