My problem is that i need this woking in the next 2,3 days. While you work on that service, how do i solve my problem using mi code ASAP?
Sorry and thank you 2011/9/15 Massimo Di Pierro <[email protected]> > You should not pass a file using xmlrpc. It is possible but not > efficient. Use a different service to upload files. I have some code > for this but need to clean it up. > > On Sep 15, 12:12 pm, António Ramos <[email protected]> wrote: > > hello, > > i receive a file via xmlrpc > > this is my function > > > > @service.xmlrpc > > def checkflow(file,name): > > > db.image.insert(file=db.image.file.store(cStringIO.StringIO(file.data),name > )) > > return "ok" > > > > It is working as expected. > > > > Now i want to open the file and process every line > > > > how do i code the opening of the file ? > > > > thank you >
