En Mon, 11 Jun 2007 17:11:23 -0300, Radamand <[EMAIL PROTECTED]> escribió:
> On Jun 11, 1:23 pm, Fredrik Lundh <[EMAIL PROTECTED]> wrote: >> here's a simple variation of that, which is a bit more efficient, and >> perhaps also a bit easier to use in the general case: >> >> while serverlist: >> still_active = [] >> for server in serverlist: >> pinger = ping[server] >> if pinger.returncode is None: >> pinger.poll() >> still_active.append(server) >> else: >> pingresult[server] = pinger.stdout.read() >> pingreturncode[server] = pinger.returncode >> serverlist = still_active >> >> </F> > > Thats an interesting approach but, if the returncode for a given > server is None say, 20 times in a row you will have append'ed that > server to the list 20 times, i suppose you could check the list to see > if its already there but thats a bit kludgey... Read the code again and notice that there are TWO lists involved. There is at most one append per server - unless there are duplicates in the original list, the new list won't have any. > also, the line "pinger = ping[server]" would have to be extracted from > this loop otherwise your going to ping the same server repeatedly > until it answers... Uh...? -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
