Elias, The resulting strategy of allow_mult=false and last_write_wins=false (which is a simplification for developer-friendliness mostly):
1) Resolve differences using the vector clock first. 2) If siblings still exist, return the one with the latest timestamp. So in a sense, it's a combination of vector-clock resolution and resolution by timestamp. This might be ok if your write rate is small, meaning that writes are unlikely to conflict; the upshot is that you don't have to worry about resolution. However, in general we suggest developers think about resolution strategies during development and then use allow_mult=true in production. On Thu, Nov 8, 2012 at 10:09 AM, Elias Levy <fearsome.lucid...@gmail.com> wrote: > On Tue, Nov 6, 2012 at 9:57 PM, Elias Levy <fearsome.lucid...@gmail.com> > wrote: >> >> It's also not clear from the docs what Riak considers the latest value >> to return if allow_mult is false and so is last_write_wins, when you >> have a conflict. > > > Any Basho folks have an answer to this one? How does Riak resolve a conflict > on a object fetch when both allow_mult and last_write_wins are false (the > default)? > > Elias > > _______________________________________________ > riak-users mailing list > riak-users@lists.basho.com > http://lists.basho.com/mailman/listinfo/riak-users_lists.basho.com > -- Sean Cribbs <s...@basho.com> Software Engineer Basho Technologies, Inc. http://basho.com/ _______________________________________________ riak-users mailing list riak-users@lists.basho.com http://lists.basho.com/mailman/listinfo/riak-users_lists.basho.com
