人言落日是天涯，望极天涯不见家 schrieb: > On Apr 29, 7:37 pm, "Daniel Nogradi" <[EMAIL PROTECTED]> 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.
You can use cStringIO for that as well. Read the module docs for it. Diez -- http://mail.python.org/mailman/listinfo/python-list
