I have only one file to process per day, so its not that important the eficiency of the code.
Em 15 de setembro de 2011 20:44, António Ramos <[email protected]>escreveu: > 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 >> > >
