if i remove the global i get an undefined error! it was suggested as a solution in this group below
http://groups.google.com/group/comp.lang.python/browse_thread/thread/b7b2dcdc3e109f3e?hide_quotes=no#msg_81305eeee43f82c6 thanks Simon Forman wrote: > a wrote: > > hi simon thanks for your reply > > You're most welcome > > > > what if i want to do this > > feed_list=[] > > feed_id=[] > > for ix in feeds_list_select: > > global feeds_list > > global feeds_id > > feeds_list.append(ix.url) > > feeds_id.append(ix.id) > > > > ie not one variable but more variables > > thanks > > in a case like this I would usually reach for the zip() function, with > the "varargs" * calling pattern > > N = [(ix.url, ix.id) for ix in feeds_list_select] > > feed_list, feed_id = zip(*N) > > > or just > > feed_list, feed_id = zip(*[(ix.url, ix.id) for ix in > feeds_list_select]) > > > btw, please note that the global statements in your example are > unnecessary.. *totally* unnecessary. :-D -- http://mail.python.org/mailman/listinfo/python-list
