If uploadseparate=True then you have to reconstruct the full path with a routine like this:
def get_path(filename,folder='uploads'): """Get full path to file stored in uploads""" f=filename.split('.') path=os.path.join(request.folder,folder, '%s.%s'%(f[0],f[1]), '%c%c'%(f[2][0],f[2][1]), filename) return path where filename passed to get_path is db.mytable[row].file_fieldname On Dec 12, 1:07 pm, "G. Clifford Williams" <g...@notadiscussion.com> wrote: > Is ther a prescribed mechanism for getting the full path to uploaded files? > > I've got an onaccept function that contains: > os.symlink(os.path.join(request.folder,'uploads', > form.vars.file_newfilename), link_path) > > which worked until I'd changed the field to use 'uploadseparate=True' > > Thanks in advance > --Cliff