> 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
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