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
