You can't offload the upload to a task in a separate process: there would be no socket to munge the data in :D Use modern js solutions to return information to the user for multiple uploads and a custom action.... and use a real webserver in front of web2py (read: nginx) that will buffer the input before passing it to web2py.
On Thursday, August 25, 2016 at 12:51:04 PM UTC+2, Mirek Zvolský wrote: > > I heard much about timeout of web request (like 60s) and about the need to > use scheduler if the action is slow. > But I'am not sure about large file uploads. > > Is the file send to the server during http request or when I ask .read() > (or .value) in the controller? Should I take care about time of the action? > And: how to give info to the user about the running upload? > > 1) > What if I use standard Web2py upload fields in forms? > > 2) > What if I use <input name="fldname" multiple>? > I think there is no built in support for this in Web2py, so I have this in > view and handle it in the controller so: > if request.vars.fldname: > for f in request.vars.fldname: > fn = f.filename.lower() > with open(os.path.join(uploadfolder, fn), 'w') as fw: > fw.write(f.value) > -- Resources: - http://web2py.com - http://web2py.com/book (Documentation) - http://github.com/web2py/web2py (Source code) - https://code.google.com/p/web2py/issues/list (Report Issues) --- You received this message because you are subscribed to the Google Groups "web2py-users" group. To unsubscribe from this group and stop receiving emails from it, send an email to web2py+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
