Michael,
Is there any chance we might see support for *with_labels* in this use
case? Even though my "solution" works, it's somewhat of an annoyance with
all these warnings being spit out:
*SAWarning: Column 'id' on table <sqlalchemy.sql.selectable.Select at
> 0x110cd5d30; Select object> being replaced by Column('id', Integer(),
> table=<Select object>, primary_key=True, nullable=False), which has the
> same key. Consider use_labels for select() statements.*
On Friday, April 10, 2015 at 12:19:31 AM UTC+2, Michael Bayer wrote:
>
>
>
> On 4/9/15 1:50 PM, Jacob Magnusson wrote:
>
> I have this case with a bundle that looks something like this:
>
> primate_bundle = Bundle(
> 'primate',
> Primate.name,
> Bundle('wooden_tool', *[
> WoodenTool.id,
> WoodenTool.name,
> WoodenToolCategory.name.label('category'),
> ]),
> Bundle('solid_tool', *[
> SolidTool.id,
> SolidTool.name,
> SolidToolCategory.name.label('category'),
> ])
> )
>
> Then I query it like this:
>
> session.query(primate_bundle)
> .select_from(Primate)
> .join(WoodenTool, Primate.main_tool)
> .join(WoodenToolCategory, WoodenTool.category_id == WoodenToolCategory.id)
> .join(SolidTool, Primate.secondary_tool)
> .join(SolidToolCategory, SolidTool.category_id == SolidToolCategory.id)
> .all()
>
> However, since the label for category name is the same within both
> sub-bundles it will throw Ambiguous column name (because the compiled SQL
> labels will be exactly the same). Adding .with_labels() doesn’t fix it.
> Full traceback can be seen by running the included examples. Commenting out
> one of the .label() lines in the example makes it runnable. Do you guys
> have a clean solution to support this use case? I really like this feature
> of creating your own custom made results so it would be a shame to not be
> able to do this.
>
> OK well here's one hacky way, that we could better accommodate by adding a
> "key" parameter to label(), this is sort of like your other approach:
>
> ay = A.y.label(None)
> ay.key = 'foobar'
> bp = B.p.label(None)
> bp.key = 'foobar'
>
> ab = Bundle(
> 'ab',
> Bundle('a', *[
> A.id,
> A.x,
> ay,
> ]),
> Bundle('b', *[
> B.id,
> B.q,
> bp,
> ])
> )
>
> The other way is to provide create_row_processor() as we describe here:
> http://docs.sqlalchemy.org/en/rel_0_9/orm/loading_columns.html?highlight=bundle#column-bundles.
>
> That way you could just make this work any way you wanted:
>
>
> from sqlalchemy import util
>
>
> class LabelBundle(Bundle):
> def __init__(self, *arg, **kw):
> self.extra_labels = kw.pop('extra_labels', {})
> super(LabelBundle, self).__init__(*arg, **kw)
>
> def create_row_processor(self, query, procs, labels):
> # or use a NamedTuple here
> keyed_tuple = util.lightweight_named_tuple(
> 'result', [self.extra_labels.get(l, l) for l in labels])
>
> def proc(row):
> return keyed_tuple([proc(row) for proc in procs])
> return proc
>
>
> ab = LabelBundle(
> 'ab',
> LabelBundle(
> 'a',
> extra_labels={'y': 'foobar'},
> *[A.id, A.x, A.y]),
> LabelBundle(
> 'b',
> extra_labels={'p': 'foobar'},
> *[B.id, B.q, B.p])
> )
>
>
>
>
>
>
>
> Tested on SQLAlchemy 1.0.0b5 and 0.9.9. Python 3.
>
> Thank you so much for any potential help you can give me on this. I’ve
> followed the source code for Bundle but I can’t think of a clean way to
> this…
>
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