Very good, thanks. Although, I'm pretty sure I understand what you are saying, what exactly do you mean by "pending/transients"?
On Feb 10, 4:13 pm, Michael Bayer <mike...@zzzcomputing.com> wrote: > 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 > > 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.
