On 2/07/2006 6:43 AM, a wrote: > if i remove the global > i get an undefined error!
That's because you have feed_list in one place and feeds_list in another. Ditto feed_id and feeds_id. Your code was *already* broken before you removed the global(s). > it was suggested as a solution in this group below Bruno wrote, in the thread that you quote: "Globals are *evil*. And functions modifying globals is the worst possible thing." Is that what you call "suggested as a solution"? > > 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
