On Feb 10, 2010, at 3:52 PM, Kent wrote: > If I understand you correctly, you are saying > object.list[0] will always cause creation (or fetch) of merged.list[0] > object.list[1] will always cause creation (or fetch) of merged.list[1] > etc. > > There may be also more merged.list[2], [3], etc... > > Correct? > > This is the merge code 0.5.8: > > if self.uselist: > dest_list = [] > for current in instances: > _recursive[(current, self)] = True > obj = session._merge(current, dont_load=dont_load, > _recursive=_recursive) > if obj is not None: > dest_list.append(obj) > if dont_load: > coll = attributes.init_collection(dest_state, > self.key) > for c in dest_list: > coll.append_without_event(c) > else: > getattr(dest.__class__, > self.key).impl._set_iterable(dest_state, dest_dict, dest_list) > > > Can I rely this implementation remaining ordered (deterministic), even > if it is re-written for optimization purposes or something?
as long as you're using lists for your relations' collection implementations there's no reason the order of pending/transients would change. The objects coming back from the DB are not deterministic unless you add order_by to your relation, but thats why i said process those separately. > > Also, I see that if obj is None, then dest_list.append() won't be > called, which would mess up my indexes. I am wondering is there a > more sure mechanism? Under what circumstances will obj be None? There's no codepath I can see where that can be None and there's no test that generates a None at that point, I'm not really sure why that check is there. I'd want to dig back to find its origins before removing it but _merge() pretty explicitly doesn't return None these days. > > > > > On Feb 10, 3:30 pm, Michael Bayer <mike...@zzzcomputing.com> wrote: >> On Feb 10, 2010, at 2:49 PM, Kent wrote: >> >> >> >>> After merge() returns, is there a way for me to pair each object in >>> the returned merge_obj with the object it was created from? >> >>> For example: >>> merged_obj = session.merge(object) >> >>> At the top level, it is trivial, merged_obj was created because of the >>> instance "object" >> >>> For single RelationProperties under the top level, it is fairly >>> simple, too. >> >>> That is: >> >>> merged.childattr was merged from object.childattr >> >>> Where it falls apart I think is if the RelationProperty.use_list == >>> True >> >>> merged.list came from object.list, but is there a way for me to >>> reference the original objects inside the list. >> >>> Did merged.list[0] come from object.list[0] or object.list[1] or >>> object_list[2]? >> >>> I particularly can't use the pk because it won't always be set (often >>> this will be a new record) >> >>> Any suggestions? >> >> the ordering of those lists (assuming they are lists and not sets) are >> deterministic, especially with regards to the pending objects that have been >> added as a result of your merge (i.e. the ones that wont have complete >> primary keys). I would match them up based on comparison of the list of >> instances that are transient/pending. >> >> >> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "sqlalchemy" group. >>> To post to this group, send email to sqlalch...@googlegroups.com. >>> To unsubscribe from this group, send email to >>> sqlalchemy+unsubscr...@googlegroups.com. >>> For more options, visit this group >>> athttp://groups.google.com/group/sqlalchemy?hl=en. >> >> > > -- > You received this message because you are subscribed to the Google Groups > "sqlalchemy" group. > To post to this group, send email to sqlalch...@googlegroups.com. > To unsubscribe from this group, send email to > sqlalchemy+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/sqlalchemy?hl=en. > -- You received this message because you are subscribed to the Google Groups "sqlalchemy" group. To post to this group, send email to sqlalch...@googlegroups.com. To unsubscribe from this group, send email to sqlalchemy+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/sqlalchemy?hl=en.
