I forgot to say, uploading an image works, if the folder isn't there it is being created, otherwise the image is uploaded to the folder.
Best, Annet Op maandag 2 november 2020 om 16:17:48 UTC+1 schreef Annet: > > To upload images to separate folders outside web2py I set the upload > folder with the model definition: > > UPLOADSDIR = '/home/me/apps/static/' > > db.define_table('mytable', > Field('image', 'upload', autodelete=True, uploadfolder=os.path.join( > UPLOADSDIR, 'vertexID%s' % session.vertexID), > ...) > > > I use the standard download function: > > def download(): > return response.download(request, db) > > > The form: > > form = SQLFORM(table, row, deletable=True, upload=URL('default', > 'download'), showid=False, buttons=buttons) > if hasattr(request.vars.image, 'filename'): > form.vars.imagefilename = request.vars.image.filename > > > and to process the form: > > if form.process().accepted: > if row and form.vars.image__delete: > file = row.image > os.remove(os.path.join(UPLOADSDIR, 'vertex%s' % > session.back_end_vertexid, file)) > > > In web2py 2.17.2 this all worked without problem, however, after upgrading > to > web2py 2.20 it no longer works. The image doesn't show in the form, it's > marked > with a broken link icon, and when I click the file link I get a page not > found error. > Also when I upload a new image, the old one isn't being deleted, so the > images pile > up in the folder. > > I hope someone knows how to solve this issue. > > > Kind regards, > > Annet > -- 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. To view this discussion on the web visit https://groups.google.com/d/msgid/web2py/cfcae028-a4fb-46ca-91b5-49abbc30a21en%40googlegroups.com.