> > > 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 -- http://mail.python.org/mailman/listinfo/python-list
