Could zipfile module process the zip data in memory?

[Daniel Nogradi]

> > > 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.

Well, as you correctly pointed out in your original post ZipFile can
receive a filename or a file-like object. If the zip archive data is
in zipFileData then you might do:

from StringIO import StringIO
from zipfile import ZipFile

data = StringIO( )
data.write( zipFileData )
data.close( )

zipp = ZipFile( data )
.........


and continue in the same way as before.

Daniel